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 53 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. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.White phosphorous reacts with aqueous NaOH solution to form $\mathrm{PH_3(g)}$ and aqueous sodium hypophosphite. 124 g of white phosphorous reacts completely with 1 L of $x$ M NaOH to form three moles of sodium hypophosphite. What is the volume (in L) of $x$ M NaOH that completely reacts with 128 g of rhombic sulphur to form aqueous sulphide, water and aqueous thiosulphate? (P= 31 u; S= 32 u) (A) 2 (B) 0.5 (C) 1 (D) 4

   ›Reveal solutionSolution

   First find $x$ from the phosphorus disproportionation stoichiometry ($x=3$ M), then apply the analogous sulfur disproportionation reaction with hot NaOH to find the NaOH needed for 128 g of rhombic sulfur — giving 2 L.

   Concept and Intuition

   Both white phosphorus and rhombic sulfur undergo base-induced disproportionation with hot concentrated NaOH: the element simultaneously oxidizes to an oxyanion and reduces to a hydride/anion. For P$_4$, this is the classic reaction producing phosphine (PH$_3$) and sodium hypophosphite (NaH$_2$PO$_2$). For S$_8$, the analogous reaction produces sodium sulfide (Na$_2$S) and sodium thiosulfate (Na$_2$S$_2$O$_3$). Solving requires first pinning down the NaOH concentration $x$ from the phosphorus data, then reusing that same concentration for the sulfur calculation.

   Step-by-Step Solution

   1. Phosphorus reaction: $\mathrm{P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2}$. Moles of P$_4$ $= \dfrac{124\ \text{g}}{124\ \text{g/mol}} = 1$ mol (since $M(\mathrm{P_4})=4\times31=124$). By stoichiometry, 1 mol P$_4$ produces exactly 3 mol NaH$_2$PO$_2$ — matching the given "3 moles of sodium hypophosphite" exactly, and consumes 3 mol NaOH. Since this used up all of 1 L of $x$ M NaOH: $x \times 1 = 3 \Rightarrow x = 3$ M.
   2. Sulfur reaction: hot concentrated NaOH disproportionates S$_8$ analogously: $\mathrm{S_8 + 12NaOH \to 4Na_2S + 2Na_2S_2O_3 + 6H_2O}$ …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.5.4 g of a metal (M) reacts with chlorine to form 26.7 g of metal chloride. What is the weight (in g) of M that reacts with 48 g of oxygen? (A) 67.5 (B) 40.5 (C) 54 (D) 27

   ›Reveal solutionSolution

   Find the equivalent weight of M from its chloride, then use the law of equivalent proportions (equal equivalents react) to find how much M combines with 48 g of oxygen. Answer: 54 g.

   Concept and Intuition

   The law of equivalent proportions says that the mass of any element combining with a fixed number of equivalents of another element is proportional to its own equivalent weight — regardless of which second element it's reacting with. So once we know M's equivalent weight from its reaction with chlorine, we can predict how much M reacts with any given mass of oxygen, without needing to know the actual chemical formula.

   Step-by-Step Solution

   1. Metal chloride formed: 26.7 g total, containing 5.4 g of M, so mass of Cl $=26.7-5.4=21.3$ g.
   2. Equivalent weight of Cl $=35.5$ g/equivalent. Equivalents of Cl reacted $=\dfrac{21.3}{35.5}=0.6$.
   3. Since equivalents of M = equivalents of Cl (law of equivalent proportions): equivalents of M $=0.6$, so equivalent weight of M $=\dfrac{5.4}{0.6}=9$ g/equivalent. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.White phosphorus reacts with aqueous NaOH to form $PH_3(g)$ and sodium hypophosphite. When 6.2g of white phosphorus reacted with 500 mL of xM NaOH solution, the concentration of sodium hypophosphite in the resultant solution was 0.3 mol $L^{-1}$. What are x (in M) and weight (in g) of $PH_3$ formed respectively? (P = 31 u; H = 1 u; O = 16 u) (A) 0.6, 1.7 (B) 0.3, 3.4 (C) 0.3, 1.7 (D) 0.6, 3.4

   ›Reveal solutionSolution

   This is the classic disproportionation of white phosphorus in hot NaOH; stoichiometry from the balanced equation gives $x=0.3\,M$ and $1.7\,g$ of $PH_3$.

   Concept and Intuition

   White phosphorus ($P_4$) undergoes disproportionation with hot concentrated alkali: some P atoms are reduced to $P^{-3}$ (as $PH_3$) while others are oxidised to $P^{+1}$ (as hypophosphite, $H_2PO_2^-$). Getting the balanced equation right lets you connect moles of $P_4$ reacted to moles of each product.

   Step-by-Step Solution

   1. Balanced equation: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$ (check: 4 P both sides; 3 Na both sides; O: $3+3=6$ left, $3\times2=6$ right; H: $3+6=9$ left, $3+6=9$ right — balanced).
   2. Moles of $P_4$ reacted $= \dfrac{6.2\,g}{4\times31\,g/mol} = \dfrac{6.2}{124} = 0.05$ mol.
   3. From stoichiometry, $0.05$ mol $P_4$ produces $0.05$ mol $PH_3$ and $3\times0.05 = 0.15$ mol $NaH_2PO_2$, consuming $3\times0.05=0.15$ mol $NaOH$. …

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7. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.The mass of ammonia (in kg) produced if 2 kg of dinitrogen reacts with 1 kg of dihydrogen is approximately (A) 4.86 (B) 2.43 (C) 3.63 (D) 6.36

   ›Reveal solutionSolution

   A limiting-reagent problem: $N_2$ runs out first, and the $NH_3$ formed is calculated from moles of $N_2$.

   Concept and Intuition

   When two reactants are given in a fixed stoichiometric reaction, whichever reactant produces the smaller amount of product when fully consumed is the limiting reagent — the actual yield must be based on it, not on whichever reactant happens to have more mass.

   Step-by-Step Solution

   1. Reaction: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
   2. Moles of $N_2 = \dfrac{2000\ \text{g}}{28\ \text{g/mol}} = 71.43$ mol.
   3. Moles of $H_2 = \dfrac{1000\ \text{g}}{2\ \text{g/mol}} = 500$ mol.
   4. $H_2$ needed to react completely with all $N_2$: $3\times 71.43 = 214.3$ mol — far less than the 500 mol available, so $N_2$ is limiting.
   5. Moles $NH_3$ produced $= 2\times 71.43 = 142.86$ mol. …

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8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.One mole of $O_2(g)$ was passed over hot coke. At the end of the reaction, 40% of $O_2(g)$ was unreacted. What is the volume (in L) of reaction mixture at STP (273.15 K and 1 bar)? (Assume only $CO(g)$ is formed in the reaction) (A) 22.7 (B) 72.64 (C) 36.32 (D) 45.4

   ›Reveal solutionSolution

   This tests stoichiometry of $2C+O_2\rightarrow 2CO$ combined with the modern IUPAC STP molar volume (22.7 L/mol, not the old 22.4 L/mol). The answer is 36.32 L.

   Concept and Intuition

   Hot coke (carbon) reacting with a limited/controlled amount of oxygen gives carbon monoxide rather than carbon dioxide when oxygen is deficient or the temperature favours CO formation; the problem tells us to assume only CO is formed. The key subtlety is that 'STP' here is explicitly defined as 273.15 K and 1 bar — this is the current IUPAC STP, whose molar gas volume is 22.7 L/mol, distinct from the older convention (1 atm, 22.4 L/mol) still used casually in many textbooks.

   Step-by-Step Solution

   1. Write the balanced reaction: $2C(s) + O_2(g) \rightarrow 2CO(g)$.
   2. Initial $O_2$ = 1 mol. 40% unreacted $\Rightarrow$ unreacted $O_2 = 0.4$ mol, so reacted $O_2 = 0.6$ mol.
   3. From stoichiometry, moles of CO formed $= 2\times(\text{mol }O_2\text{ reacted}) = 2\times0.6 = 1.2$ mol.
   4. Total gaseous moles in the final mixture $=$ unreacted $O_2$ + CO formed $= 0.4 + 1.2 = 1.6$ mol. …

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9. COMEDK 2026Set 2026-A1 markMCQ
   Q.How many molecules of $\mathrm{CO}_2(\mathrm{~g})$ are obtained on reaction of 24 grams of methane with 4 moles of oxygen? (A) $3.011 \times 10^{23}$ (B) $12.044 \times 10^{23}$ (C) $6.022 \times 10^{23}$ (D) $9.033 \times 10^{23}$

   ›Reveal solutionSolution

   Comparing the mole ratio of methane to oxygen against the stoichiometric requirement shows methane is the limiting reagent, producing 1.5 mol of CO$_2$ — $9.033\times10^{23}$ molecules, option (D).

   Step-by-step reasoning

   1. Balanced equation.

   $\mathrm{CH_4 + 2O_2 \to CO_2 + 2H_2O}$

   2. Convert to moles.

      Molar mass of CH$_4$ = 16 g/mol, so moles of CH$_4$ = $24/16 = 1.5$ mol. Moles of O$_2$ = 4 mol (given).

   3. Find the limiting reagent.

      1.5 mol CH$_4$ requires $1.5 \times 2 = 3$ mol O$_2$. We have 4 mol O$_2$ available — more than enough — so methane is limiting.

   4. Moles of CO$_2$ produced.

      1 mol CH$_4$ gives 1 mol CO$_2$, so 1.5 mol CH$_4$ gives 1.5 mol CO$_2$.

   5. Convert to molecules.

      $1.5 \times 6.022\times10^{23} = 9.033\times10^{23} \text{ molecules}$ …

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10. COMEDK 2026Set 2026-M1 markMCQ
    Q.The mass of precipitate formed when 50 mL of $16.9 \%$ aqueous solution of $\mathrm{AgNO}_3$ is mixed with 50 mL of $5.8 \% \mathrm{NaCl}$ solution is-----------$[\mathrm{Ag}=107.8, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{Cl}=35.5]$ (A) 14 (B) 5 (C) 7 (D) 6

    ›Reveal solutionSolution

    The key is to identify the limiting reagent in the precipitation reaction $\mathrm{AgNO_3 + NaCl \to AgCl \downarrow + NaNO_3}$ by converting the given mass percentages to moles, then calculating the mass of $\mathrm{AgCl}$ formed. The precipitate mass is $7$ g, so the correct option is (C).

    ---

    Concept and Intuition

    When two solutions are mixed, a precipitation reaction occurs if the product is insoluble. Here, silver nitrate and sodium chloride react to form silver chloride, which is a white solid. The mass of precipitate depends entirely on the limiting reagent — the reactant that runs out first. We are given mass percentages, not molarities, so we must first find the actual mass of each solute in the mixed volume, then convert to moles, and finally see which one limits the reaction.

    A common mistake is to assume equal volumes mean equal moles, but the concentrations (mass percentages) are different, so the mole amounts differ. Always check the stoichiometry.

    ---

    Step-by-step solution

    1. Find the mass of each solute in the given volumes.

    We have 50 mL of each solution. The density of dilute aqueous solutions is approximately $1\ \mathrm{g/mL}$, so 50 mL ≈ 50 g of solution.

    • For $\mathrm{AgNO_3}$:

      Mass of solution = 50 g

      Percentage = $16.9\%$

      Mass of $\mathrm{AgNO_3}$ = $\frac{16.9}{100} \times 50 = 8.45\ \mathrm{g}$

    • For $\mathrm{NaCl}$:

      Mass of solution = 50 g

      Percentage = $5.8\%$

      Mass of $\mathrm{NaCl}$ = $\frac{5.8}{100} \times 50 = 2.9\ \mathrm{g}$

    2. Convert these masses to moles.

    Molar mass of $\mathrm{AgNO_3}$:

    $\mathrm{Ag} = 107.8,\ \mathrm{N} = 14,\ \mathrm{O_3} = 48$

    Total = $107.8 + 14 + 48 = 169.8\ \mathrm{g/mol}$

    Moles of $\mathrm{AgNO_3}$ = $\frac{8.45}{169.8} \approx 0.04976\ \mathrm{mol}$

    Molar mass of $\mathrm{NaCl}$:

    $\mathrm{Na} = 23,\ \mathrm{Cl} = 35.5$

    Total = $23 + 35.5 = 58.5\ \mathrm{g/mol}$

    Moles of $\mathrm{NaCl}$ = $\frac{2.9}{58.5} \approx 0.04957\ \mathrm{mol}$

    3. Write the balanced reaction and identify the limiting reagent.

    $\mathrm{AgNO_3 + NaCl \to AgCl \downarrow + NaNO_3}$

    The mole ratio is 1:1.

    We have:

    • $\mathrm{AgNO_3}$: $0.04976\ \mathrm{mol}$
    • $\mathrm{NaCl}$: $0.04957\ \mathrm{mol}$ …

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11. MHT-CET 2026Set pcm-2026-04-15-M1 markMCQ
    Q.What is the approximate mass of the precipitate formed when 50 mL of 16.9% solution of 3 is mixed with 50 mL of 7.45% KCl solution? (Molar mass of 3=169 g/mol, KCl =74.5 g/mol, AgCl =143.3 g/mol) (A) 3.5 g (B) 7 g (C) 14 g (D) 28 g

    ›Reveal solutionSolution

    Compute moles of AgNO₃ and KCl from the percentage solutions, then find the limiting AgCl precipitate mass.

    50 mL of 16.9% AgNO₃ contains 8.45 g → 8.45/169 = 0.05 mol. 50 mL of 7.45% KCl contains 3.725 g → 3.725/74.5 = 0.05 mol. AgNO₃ + KCl → AgCl + KNO₃, 1:1 stoi …

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12. 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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