Q.In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Noble metals like Au, Pt dissolve in a mixture of (A) 1 part of Conc. HCl + 1 part of Conc. HNO$_3$ (B) 1 part of Conc. H$_2$SO$_4$ + 1 part of Conc. HNO$_3$ (C) 1 part of Conc. HNO$_3$ + 3 parts of Conc. HCl (D) 3 parts of Conc. HNO$_3$ + 1 part of Conc. HCl

   ›Reveal solutionSolution

   Noble metals like gold and platinum are famously inert, but they dissolve in aqua regia — a specific mixture of concentrated nitric acid and hydrochloric acid in a 1:3 volume ratio. The correct option is (C).

   The key concept here is chemical reactivity and complex formation. Gold and platinum are "noble" because they resist oxidation by most single acids. However, a mixture of HCl and HNO₃ works through a clever synergy: nitric acid oxidizes the metal, and the chloride ions from HCl then complex the oxidized metal ions, pulling them into solution and preventing re-deposition. The classic recipe is 1 part concentrated HNO₃ to 3 parts concentrated HCl — this is aqua regia.

   Let’s break it down:

   1. Why not a single acid?

      Gold (Au) and platinum (Pt) have high reduction potentials. For example, Au³⁺ + 3e⁻ → Au has $E^\circ = +1.50\ \text{V}$. Nitric acid alone ($E^\circ \approx +0.96\ \text{V}$ for NO₃⁻/NO) cannot oxidize gold. Hydrochloric acid alone cannot either — it’s a non-oxidizing acid.

   2. The synergy in aqua regia:

      In the mixture, nitric acid acts as the oxidizer:

   $\text{Au} + 3\text{NO}_3^- + 6\text{H}^+ \rightarrow \text{Au}^{3+} + 3\text{NO}_2 + 3\text{H}_2\text{O}$

   But the Au³⁺ ions would quickly be reduced back by chloride ions if not stabilized. However, chloride ions from HCl form a very stable complex:

   $\text{Au}^{3+} + 4\text{Cl}^- \rightarrow [\text{AuCl}_4]^-$

   This tetrachloroaurate(III) complex is highly soluble and shifts the equilibrium, allowing the oxidation to proceed.

   3. The exact ratio matters: …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In the estimation of nitrogen by Kjeldahl’s method, the ammonia produced by 2.0 g of an organic compound was completely neutralized by 20 mL of 2M sulphuric acid. The percentage of nitrogen in the compound is (A) 56 (B) 28 (C) 36 (D) 46

   ›Reveal solutionSolution

   In Kjeldahl’s method, the ammonia from the sample is absorbed in excess acid, and the unused acid is back-titrated. Here, the ammonia itself neutralizes the entire 20 mL of 2M $\ce{H2SO4}$, so the moles of nitrogen equal twice the moles of acid. The percentage of nitrogen is 56%.

   The heart of Kjeldahl’s method is converting the nitrogen in an organic compound into ammonia ($\ce{NH3}$), then trapping that ammonia in a known volume of standard acid. The amount of acid that gets neutralized tells you how much ammonia was produced — and from that, the nitrogen content.

   In this problem, the ammonia from 2.0 g of sample completely neutralized 20 mL of 2M sulphuric acid. That means all the acid was used up; there’s no leftover to back-titrate. So the moles of acid consumed equal the moles of $\ce{H2SO4}$ originally taken.

   Let’s walk through the stoichiometry carefully.

   1. Moles of $\ce{H2SO4}$ used

      Volume = 20 mL = 0.020 L, concentration = 2 mol/L.

      Moles of $\ce{H2SO4} = 2 \times 0.020 = 0.040$ mol.

   2. Reaction between ammonia and sulphuric acid

      The neutralization is:

   $2\ce{NH3} + \ce{H2SO4} \rightarrow (\ce{NH4})2\ce{SO4}$

   So 1 mole of $\ce{H2SO4}$ reacts with 2 moles of $\ce{NH3}$.

   Therefore, moles of $\ce{NH3}$ produced = $2 \times 0.040 = 0.080$ mol.

   3. Moles of nitrogen

      Each $\ce{NH3}$ molecule contains one nitrogen atom. So moles of nitrogen = moles of $\ce{NH3}$ = 0.080 mol.

   4. Mass of nitrogen

      Atomic mass of N = 14 g/mol.

      Mass of nitrogen = $0.080 \times 14 = 1.12$ g.

   5. Percentage of nitrogen

      Sample mass = 2.0 g.

      Percentage = $\frac{1.12}{2.0} \times 100 = 56\%$. …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The reactions which produce O$_2$ are I. \quad 2 KClO$_3$ \xrightarrow{\Delta \ MnO_2} II. \quad 2 KMnO$_4$ \xrightarrow{\Delta} III. \quad (NH$_4$)$_2$Cr$_2$O$_7$ \xrightarrow{\Delta} The correct answer is (A) I, III only (B) I, II, III (C) II, III only (D) I, II only

   ›Reveal solutionSolution

   We need to identify which of the given thermal decomposition reactions produce oxygen gas ($O_2$). Reactions I ($2 \text{KClO}_3 \xrightarrow{\Delta \ \text{MnO}_2}$) and II ($2 \text{KMnO}_4 \xrightarrow{\Delta}$) produce $O_2$, while Reaction III ($(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta}$) produces nitrogen gas ($N_2$). The correct option is (D).

   The problem asks us to identify which of the given reactions produce oxygen gas ($O_2$). All three reactions are examples of thermal decomposition, where a compound breaks down into simpler substances upon heating. To solve this, we need to recall the specific products formed during the thermal decomposition of each reactant.

   1. Analyze Reaction I: $2 \text{KClO}_3 \xrightarrow{\Delta \ \text{MnO}_2}$ This reaction involves the thermal decomposition of potassium chlorate ($\text{KClO}_3$). Potassium chlorate decomposes upon heating to form potassium chloride ($\text{KCl}$) and oxygen gas ($\text{O}_2$). Manganese dioxide ($\text{MnO}_2$) acts as a catalyst, lowering the activation energy and allowing the decomposition to occur at a lower temperature and faster rate, but it does not participate in the reaction itself. The balanced chemical equation is:

   $2 \text{KClO}_3(s) \xrightarrow{\Delta \ \text{MnO}_2} 2 \text{KCl}(s) + 3 \text{O}_2(g)$

   Since oxygen gas is produced, Reaction I is one of the correct options.
   

   2. Analyze Reaction II: $2 \text{KMnO}_4 \xrightarrow{\Delta}$

   This reaction involves the thermal decomposition of potassium permanganate ($\text{KMnO}_4$). When heated, potassium permanganate decomposes to form potassium manganate ($\text{K}_2\text{MnO}_4$), manganese dioxide ($\text{MnO}_2$), and oxygen gas ($\text{O}_2$).

   The balanced chemical equation is:

   $2 \text{KMnO}_4(s) \xrightarrow{\Delta} \text{K}_2\text{MnO}_4(s) + \text{MnO}_2(s) + \text{O}_2(g)$

   Since oxygen gas is produced, Reaction II is also one of the correct options.
   

   3. Analyze Reaction III: $(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta}$ …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.In which of the following, oxidation state of nitrogen is lowest? (A) $\mathrm{NH_2OH}$ (B) $\mathrm{NH_4Cl}$ (C) $\mathrm{N_2H_4}$ (D) $\mathrm{HNO_2}$

   ›Reveal solutionSolution

   The oxidation state of nitrogen is lowest in $\mathrm{NH_4Cl}$ (ammonium chloride), where it is $-3$, because nitrogen is bonded to four hydrogen atoms (each +1) and the chloride ion is a spectator.

   The key idea is to assign oxidation numbers using the standard rules: hydrogen is usually +1 (except in metal hydrides), oxygen is usually -2 (except in peroxides), and the sum of oxidation numbers in a neutral compound is zero (or equals the charge for polyatomic ions). The lowest (most negative) oxidation state for nitrogen is the one where it is most reduced, i.e., bonded to the most hydrogen atoms.

   Let’s work through each option step by step.

   1. Option (A): $\mathrm{NH_2OH}$ (hydroxylamine)

      • Structure: $\mathrm{H_2N-OH}$.
      • Hydrogen: 3 H atoms × (+1) = +3.
      • Oxygen: 1 O atom × (-2) = -2.
      • Let nitrogen oxidation state = $x$.
      • Sum: $x + 3 - 2 = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1$.
      • So nitrogen is $-1$ here.

   2. Option (B): $\mathrm{NH_4Cl}$ (ammonium chloride)

      • This is an ionic compound: $\mathrm{NH_4^+}$ and $\mathrm{Cl^-}$.
      • In the ammonium ion, hydrogen: 4 H × (+1) = +4.
      • Let nitrogen = $x$.
      • Sum for the ion: $x + 4 = +1 \Rightarrow x = -3$.
      • So nitrogen is $-3$ here.

   3. Option (C): $\mathrm{N_2H_4}$ (hydrazine)

      • Hydrogen: 4 H × (+1) = +4.
      • Let each nitrogen = $x$ (they are equivalent).
      • Sum: $2x + 4 = 0 \Rightarrow 2x = -4 \Rightarrow x = -2$.
      • So each nitrogen is $-2$.

   4. Option (D): $\mathrm{HNO_2}$ (nitrous acid)

      • Hydrogen: +1.
      • Oxygen: 2 O × (-2) = -4.
      • Let nitrogen = $x$.
      • Sum: $1 + x - 4 = 0 \Rightarrow x - 3 = 0 \Rightarrow x = +3$. …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The number of moles of oxalate ions oxidized by one mole of permanganate ions in acidic medium is (A) $2.5$ (B) $5.0$ (C) $1.5$ (D) $2.0$

   ›Reveal solutionSolution

   In acidic medium, one mole of $\mathrm{MnO_4^-}$ gains 5 electrons, while one mole of $\mathrm{C_2O_4^{2-}}$ loses 2 electrons. By electron balance, 1 mole of $\mathrm{MnO_4^-}$ oxidizes $2.5$ moles of oxalate. The correct option is (A).

   ---

   The key is to remember that in redox titrations, the number of electrons lost by the reducing agent must equal the number of electrons gained by the oxidizing agent. Here, permanganate ion ($\mathrm{MnO_4^-}$) is the oxidizer, and oxalate ion ($\mathrm{C_2O_4^{2-}}$) is the reducer. The question asks: for every one mole of permanganate, how many moles of oxalate get oxidized? That is simply the ratio of electrons transferred, inverted.

   Why this works: Instead of memorizing a balanced equation, you can find the mole ratio directly from the change in oxidation states. This is faster and less error-prone.

   ---

   1. Find the change in oxidation state for manganese.

      In $\mathrm{MnO_4^-}$, manganese is in the +7 oxidation state (since each oxygen is -2, total -8, and the ion charge is -1, so Mn must be +7). In acidic medium, $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$, where manganese is +2.

      The change: $+7 \rightarrow +2$ means a gain of 5 electrons per Mn atom.

   2. Find the change in oxidation state for carbon in oxalate.

      Oxalate ion is $\mathrm{C_2O_4^{2-}}$. Each oxygen is -2, so total from four oxygens is -8. The ion charge is -2, so the two carbons together must have an oxidation state of +6. That means each carbon is +3.

      In acidic medium, oxalate is oxidized to carbon dioxide ($\mathrm{CO_2}$), where carbon is +4.

      The change per carbon: $+3 \rightarrow +4$ means a loss of 1 electron per carbon.

      Since each oxalate ion contains two carbons, one mole of oxalate loses 2 electrons total.

   3. Balance the electrons to find the mole ratio.

      One mole of $\mathrm{MnO_4^-}$ accepts 5 moles of electrons. …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.How many of the following metals give oxides and nitrides when burnt in air? Be, Na, Mg, Ba, Sr, Li, K (A) 5 (B) 4 (C) 2 (D) 3

   ›Reveal solutionSolution

   When burnt in air, most metals form oxides, but only highly reactive metals, particularly alkaline earth metals and lithium, also react with nitrogen to form nitrides. The metals from the given list that form both oxides and nitrides are Be, Mg, Ba, Sr, and Li, making a total of 5.

   Concept and Intuition

   Air is primarily a mixture of nitrogen (about 78%) and oxygen (about 21%). When metals are burnt in air, they readily react with oxygen to form various oxides. However, nitrogen gas ($N_2$) is very unreactive due to the strong triple bond between its two atoms. For a metal to react with nitrogen and form a nitride, it must be a strong enough reducing agent to break this triple bond.

   Generally, highly electropositive metals, especially those with a relatively high charge density (like lithium and the alkaline earth metals), are capable of reacting with nitrogen to form nitrides.

   • Alkali Metals (Group 1): Most alkali metals react vigorously with oxygen to form oxides, peroxides, or superoxides depending on the metal and conditions. However, among the alkali metals, only lithium (Li) is unique in its ability to react directly with nitrogen at high temperatures to form lithium nitride ($Li_3N$). This is attributed to the small size and high charge density of the $Li^+$ ion, which allows for a high lattice energy in $Li_3N$. Other alkali metals like sodium (Na) and potassium (K) do not react with nitrogen under normal burning conditions.
   • Alkaline Earth Metals (Group 2): All alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) are highly reactive and readily react with both oxygen to form oxides ($MO$) and nitrogen to form nitrides ($M_3N_2$) when heated in air.

   Step-by-step Analysis

   Let's examine each metal from the given list:

   1. Beryllium (Be): Beryllium is an alkaline earth metal (Group 2).

      • Reacts with oxygen to form beryllium oxide: $2Be(s) + O_2(g) \xrightarrow{\text{heat}} 2BeO(s)$
      • Reacts with nitrogen to form beryllium nitride: $3Be(s) + N_2(g) \xrightarrow{\text{heat}} Be_3N_2(s)$
      • Conclusion: Be forms both oxides and nitrides.

   2. Sodium (Na): Sodium is an alkali metal (Group 1).

      • Reacts with oxygen to primarily form sodium peroxide, with some sodium oxide and superoxide: $2Na(s) + O_2(g) \xrightarrow{\text{heat}} Na_2O_2(s)$ (main product)
      • Does not react with nitrogen under normal burning conditions.
      • Conclusion: Na forms oxides but not nitrides.

   3. Magnesium (Mg): Magnesium is an alkaline earth metal (Group 2).

      • Reacts with oxygen to form magnesium oxide: $2Mg(s) + O_2(g) \xrightarrow{\text{heat}} 2MgO(s)$
      • Reacts with nitrogen to form magnesium nitride: $3Mg(s) + N_2(g) \xrightarrow{\text{heat}} Mg_3N_2(s)$
      • Conclusion: Mg forms both oxides and nitrides.

   4. Barium (Ba): Barium is an alkaline earth metal (Group 2).

      • Reacts with oxygen to form barium oxide: $2Ba(s) + O_2(g) \xrightarrow{\text{heat}} 2BaO(s)$
      • Reacts with nitrogen to form barium nitride: $3Ba(s) + N_2(g) \xrightarrow{\text{heat}} Ba_3N_2(s)$
      • Conclusion: Ba forms both oxides and nitrides.

   5. Strontium (Sr): Strontium is an alkaline earth metal (Group 2). …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The number of moles of oxalate ions oxidized by one mole of permanganate ions in acidic medium is (A) $2.5$ (B) $2.0$ (C) $1.5$ (D) $5.0$

   ›Reveal solutionSolution

   In acidic medium, permanganate ($\text{MnO}_4^-$) is reduced to $\text{Mn}^{2+}$ (5-electron change), while oxalate ($\text{C}_2\text{O}_4^{2-}$) is oxidized to $\text{CO}_2$ (2-electron change per ion). Balancing electrons gives 2.5 moles of oxalate per mole of permanganate, so the answer is (A).

   ---

   The key here is to recognize that this is a redox stoichiometry problem. The number of moles of one species oxidized by one mole of another depends entirely on the number of electrons each gains or loses. You don't need the full balanced equation — just the half-reactions and the electron transfer.

   Why this works:

   In acidic solution, permanganate ion ($\text{MnO}_4^-$) always acts as a strong oxidizing agent, gaining 5 electrons to become $\text{Mn}^{2+}$. Oxalate ion ($\text{C}_2\text{O}_4^{2-}$) is a reducing agent; each ion loses 2 electrons to form two molecules of $\text{CO}_2$. The mole ratio is simply the ratio of electrons exchanged: 5 electrons from permanganate need to be matched by 5 electrons from oxalate, so you need $\frac{5}{2} = 2.5$ oxalate ions per permanganate ion.

   ---

   Step-by-step reasoning:

   1. Write the reduction half-reaction for permanganate in acid:

      $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

      This shows that 1 mole of $\text{MnO}_4^-$ accepts 5 moles of electrons.

   2. Write the oxidation half-reaction for oxalate:

      $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$

      This shows that 1 mole of $\text{C}_2\text{O}_4^{2-}$ donates 2 moles of electrons.

   3. Balance the electrons transferred:

      To use all 5 electrons from one permanganate, we need oxalate to supply exactly 5 electrons.

      If 1 oxalate gives 2 electrons, then the number of oxalate moles needed is: …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.12 g of an element reacts with 32 g of oxygen. What is the equivalent weight of the element? (A) $12$ (B) $6$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   The equivalent weight of an element is the mass that combines with or displaces 8 g of oxygen. Here, 12 g of element reacts with 32 g of oxygen, so the equivalent weight is (12/32) × 8 = 3 g/eq. The correct option is (D).

   The key concept is equivalent weight in the context of oxide formation. Equivalent weight is defined as the mass of an element that combines with 8 grams of oxygen (since the equivalent weight of oxygen is 8, based on its valency of 2 and atomic mass of 16). So instead of memorizing formulas, think: If 32 g of oxygen is "worth" 4 equivalents (because 32 ÷ 8 = 4), then the 12 g of element must also be 4 equivalents. Therefore, one equivalent of the element weighs 12 ÷ 4 = 3 g.

   Let’s walk through it step by step.

   1. Recall the definition of equivalent weight for oxygen.

      The equivalent weight of oxygen is always 8 g/eq because oxygen has an atomic mass of 16 and a valency of 2 (16 ÷ 2 = 8). This is a fixed reference point in equivalent weight calculations.

   2. Find how many equivalents of oxygen are present.

      We have 32 g of oxygen. Since 1 equivalent of oxygen = 8 g, the number of equivalents of oxygen is:

   $\text{Equivalents of oxygen} = \frac{32 \text{ g}}{8 \text{ g/eq}} = 4 \text{ eq}$

   3. Apply the law of equivalence.

      In a chemical reaction, the number of equivalents of the element must equal the number of equivalents of oxygen (because they combine in equivalent proportions). So the element also provides 4 equivalents.

   4. Calculate the equivalent weight of the element. …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The common components of photochemical smog are (A) $\mathrm{O_3, \ CH_4, \ CO_2}$ (B) $\mathrm{O_3, \ CO_2, \ CO}$ (C) $\mathrm{O_2, \ SO_3, \ PAN}$ (D) $\mathrm{O_3, \ NO, \ PAN}$

   ›Reveal solutionSolution

   Photochemical smog is formed from nitrogen oxides and volatile organic compounds under sunlight; its key components are ozone ($\mathrm{O_3}$), nitric oxide ($\mathrm{NO}$), and peroxyacyl nitrates (PAN). The correct choice is (D).

   Photochemical smog is a type of air pollution that forms when sunlight reacts with pollutants from vehicle exhaust and industrial emissions. The key ingredients are nitrogen oxides ($\mathrm{NO_x}$) and volatile organic compounds (VOCs). Sunlight drives a complex chain of reactions that produce secondary pollutants — most notably ground-level ozone ($\mathrm{O_3}$), nitric oxide ($\mathrm{NO}$), and peroxyacyl nitrates (PAN). Methane ($\mathrm{CH_4}$) and carbon dioxide ($\mathrm{CO_2}$) are not characteristic components; they are either a greenhouse gas or a combustion product, not a direct product of photochemical smog chemistry. Sulfur trioxide ($\mathrm{SO_3}$) is associated with industrial smog, not photochemical smog.

   Let’s examine each option:

   1. Option (A): $\mathrm{O_3, \ CH_4, \ CO_2}$

      • Ozone is indeed a major component of photochemical smog.
      • Methane ($\mathrm{CH_4}$) is a simple hydrocarbon, but it is not a typical reactive VOC in smog formation (it reacts slowly) and is not a component of the smog itself.
      • Carbon dioxide ($\mathrm{CO_2}$) is a stable end product of combustion, not a reactive smog component. → This option is incorrect.

   2. Option (B): $\mathrm{O_3, \ CO_2, \ CO}$

      • Ozone is correct.
      • Carbon dioxide and carbon monoxide are not characteristic photochemical smog components; CO is a primary pollutant from incomplete combustion, not a secondary smog product. → This option is incorrect.

   3. Option (C): $\mathrm{O_2, \ SO_3, \ PAN}$

      • Oxygen ($\mathrm{O_2}$) is normal atmospheric gas, not a pollutant.
      • Sulfur trioxide ($\mathrm{SO_3}$) is involved in acid rain and industrial smog, not photochemical smog.
      • PAN (peroxyacetyl nitrate) is a correct component. → This option is incorrect.

   4. Option (D): $\mathrm{O_3, \ NO, \ PAN}$ …

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10. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.The oxide of nitrogen, ‘X’ is a blue solid and is acidic in nature. ‘X’ is (A) $\mathrm{N_2O_4}$ (B) $\mathrm{N_2O_3}$ (C) $\mathrm{N_2O_5}$ (D) $\mathrm{NO_2}$

    ›Reveal solutionSolution

    The key is that a blue, acidic nitrogen oxide must be $\mathrm{N_2O_3}$, which forms a blue solid when cooled and reacts with water to give nitrous acid. The correct option is (B).

    The question asks us to identify a nitrogen oxide that is a blue solid and acidic in nature. Let’s think about what these clues mean.

    Concept and intuition:

    Nitrogen forms several oxides with different colors and acid-base behaviors. The color “blue” is rare among common nitrogen oxides — most are colorless, brown, or white. The only well-known blue nitrogen oxide is dinitrogen trioxide ($\mathrm{N_2O_3}$), which is a deep blue liquid or solid at low temperatures. Its acidity comes from reacting with water to form nitrous acid ($\mathrm{HNO_2}$), a weak acid. The other options either aren’t blue or aren’t solids under normal conditions.

    Let’s check each option step by step.

    1. Option (A): $\mathrm{N_2O_4}$

       • This is actually a dimer of $\mathrm{NO_2}$. At room temperature, it exists as a brown gas (mixed with $\mathrm{NO_2}$). When cooled, it forms a colorless solid (not blue).
       • It is acidic (forms nitric and nitrous acids with water), but the color doesn’t match. So (A) is out.

    2. Option (B): $\mathrm{N_2O_3}$

       • Dinitrogen trioxide is a deep blue solid or liquid at low temperatures (below about $-20^\circ\mathrm{C}$).
       • It is acidic: with water, it gives nitrous acid:

    $\mathrm{N_2O_3 + H_2O \rightarrow 2HNO_2}$

    • Both clues — blue solid and acidic — fit perfectly. This is the correct answer.

    3. Option (C): $\mathrm{N_2O_5}$
       • Dinitrogen pentoxide is a white crystalline solid (not blue). …

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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.The masses of carbondioxide and water (in g) respectively formed during complete combustion of 10 g of glucose at STP are (A) 14.66, 18.0 (B) 14.66, 6.0 (C) 12.0, 6.0 (D) 24.0, 12.0

    ›Reveal solutionSolution

    The key is to use the balanced combustion equation of glucose and stoichiometric mole ratios.

    10 g of glucose (C₆H₁₂O₆) yields 14.66 g CO₂ and 6.0 g H₂O, so the correct option is (B).

    Concept & Intuition

    Combustion of glucose is a classic stoichiometry problem. The balanced equation tells us exactly how many moles of CO₂ and H₂O are produced per mole of glucose burned. Since mass is conserved, we convert the given mass of glucose to moles, then use the mole ratios to find moles of products, and finally convert those moles back to grams. The trick is to be careful with molar masses and to remember that water is produced as a gas at STP, but we are asked for mass, not volume.

    Step-by-step solution

    1. Write the balanced combustion equation Glucose (C₆H₁₂O₆) reacts with oxygen to give carbon dioxide and water:

    $\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O}$

    This shows that 1 mole of glucose produces 6 moles of CO₂ and 6 moles of H₂O.

    2. Find the molar mass of glucose

    $\text{Molar mass of C₆H₁₂O₆} = 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 = 180.156 \ \text{g/mol}$

    For simplicity, we often use 180 g/mol, but let’s keep precision: ≈ 180.16 g/mol.

    3. Calculate moles of glucose in 10 g

    $\text{Moles of glucose} = \frac{10\ \text{g}}{180.16\ \text{g/mol}} \approx 0.0555\ \text{mol}$

    4. Use the mole ratio to find moles of CO₂ and H₂O From the equation: 1 mol glucose → 6 mol CO₂ and 6 mol H₂O.

    $\text{Moles of CO₂} = 0.0555 \times 6 = 0.333\ \text{mol}$

    $\text{Moles of H₂O} = 0.0555 \times 6 = 0.333\ \text{mol}$

    5. Convert moles to grams
       • Molar mass of CO₂ = 12.01 + 2(16.00) = 44.01 g/mol $\text{Mass of CO₂} = 0.333 \times 44.01 \approx 14.66\ \text{g}$ …

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12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.Which of the following reaction gives nitrogen (II) oxide as one of the products? (A) Cu + dil HNO$_3$ $\rightarrow$ (B) Cu + conc. HNO$_3$ $\rightarrow$ (C) Zn + dil. HNO$_3$ $\rightarrow$ (D) Zn + conc. HNO$_3$ $\rightarrow$

    ›Reveal solutionSolution

    The key is that dilute nitric acid is reduced to NO (nitrogen(II) oxide) by moderately active metals like copper, while concentrated HNO₃ gives NO₂ and zinc with dilute acid can give N₂O or NH₄⁺. The correct option is (A).

    The question asks which reaction produces nitrogen(II) oxide (NO) as a product. This is a classic problem about the reduction products of nitric acid, which depend on both the concentration of the acid and the reducing power of the metal.

    Concept & Intuition:

    Nitric acid (HNO₃) is a strong oxidizing agent. When it reacts with a metal, it is reduced — but the product varies. The key rule:

    • Concentrated HNO₃ tends to be reduced to nitrogen dioxide (NO₂, brown gas).
    • Dilute HNO₃ tends to be reduced to nitric oxide (NO, colorless gas that turns brown in air).
    • With very strong reducing agents (like zinc) and very dilute acid, further reduction to N₂O or even NH₄⁺ can occur. Copper is a moderately active metal; with dilute HNO₃ it gives NO, while with concentrated HNO₃ it gives NO₂. Zinc is more reactive and can reduce dilute HNO₃ beyond NO.

    Let’s examine each option step by step.

    1. Option (A): Cu + dil HNO₃ Copper reacts with dilute nitric acid to give copper(II) nitrate, water, and nitric oxide (NO). The balanced equation is:

    $3\text{Cu} + 8\text{HNO}_3(\text{dil}) \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} \uparrow + 4\text{H}_2\text{O}$

    Here, NO is indeed produced. This matches the question.

    2. Option (B): Cu + conc. HNO₃ With concentrated nitric acid, copper gives nitrogen dioxide instead:

    $\text{Cu} + 4\text{HNO}_3(\text{conc}) \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 \uparrow + 2\text{H}_2\text{O}$

    NO₂ is a brown gas, not NO. So this is incorrect.

    3. Option (C): Zn + dil HNO₃ Zinc is a stronger reducing agent than copper. With very dilute nitric acid, zinc can reduce the nitrate ion further — often to nitrous oxide (N₂O) or even ammonium nitrate (NH₄NO₃). For example:

    $4\text{Zn} + 10\text{HNO}_3(\text{very dil}) \rightarrow 4\text{Zn(NO}_3)_2 + \text{N}_2\text{O} \uparrow + 5\text{H}_2\text{O}$

    or with even more dilute acid:

    $$$$

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