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 19 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.An ideal gas mixture of $\mathrm{C_2H_6}$ and $\mathrm{C_2H_4}$ occupies a volume of 28 L at 1 atm and 273 K. This mixture reacts completely with 128 g of $\mathrm{O_2}$ to produce $\mathrm{CO_2}$ and $\mathrm{H_2O}(l)$. What is the mole fraction of $\mathrm{C_2H_4}$ in the mixture? (A) 0.4 (B) 0.8 (C) 0.5 (D) 0.6

   ›Reveal solutionSolution

   Use ideal gas law to get total moles, then set up simultaneous equations from the O$_2$ stoichiometry of the two combustion reactions; the answer is 0.6.

   Concept and Intuition

   At 1 atm and 273 K the molar volume is the standard 22.4 L/mol, so the total moles of the gas mixture follow directly from its volume. The O$_2$ consumed by a hydrocarbon mixture is a weighted sum of each component's own combustion stoichiometry — this lets us set up two simultaneous linear equations (total moles, total O$_2$) to solve for the two unknown mole amounts.

   Step-by-Step Solution

   1. Total moles of gas mixture: $n = \dfrac{PV}{RT}$, or simply $n = \dfrac{28\text{ L}}{22.4\text{ L/mol}} = 1.25\ \text{mol}$ (since conditions are STP).
   2. Let $a$ = mol C$_2$H$_6$ and $b$ = mol C$_2$H$_4$: $a + b = 1.25$.
   3. Combustion reactions: $\mathrm{C_2H_6 + \tfrac{7}{2}O_2 \to 2CO_2 + 3H_2O}$ $\mathrm{C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O}$
   4. Moles of O$_2$ used: $128\text{ g}/32\text{ g mol}^{-1} = 4\ \text{mol}$, so $3.5a + 3b = 4$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Complete combustion of ethane gives only gaseous products. In a closed vessel, 15 g of ethane and 112 g of $\mathrm{O_2}$ were allowed to completely react. What is the total number of moles of gaseous substances present in the vessel at the end of the reaction? (A) 4.25 (B) 2.5 (C) 1.75 (D) 8.50

   ›Reveal solutionSolution

   Ethane is the limiting reagent; after combustion the vessel holds CO$_2$, H$_2$O (vapour) and leftover O$_2$ — total 4.25 mol.

   Concept and Intuition

   With two reactants given, the first job is always to identify the limiting reagent by comparing the O$_2$ actually required against the O$_2$ available. Since the problem states combustion gives only gaseous products, water counts as a gas here (not condensed), so it must be included in the total mole count along with any unreacted O$_2$.

   Step-by-Step Solution

   1. Moles of ethane = $15/30 = 0.5\ \text{mol}$ (molar mass of C$_2$H$_6$ = 30 g/mol).
   2. Moles of O$_2$ supplied = $112/32 = 3.5\ \text{mol}$.
   3. Balanced equation: $\mathrm{C_2H_6 + \tfrac{7}{2}O_2 \to 2CO_2 + 3H_2O}$.
   4. O$_2$ required for 0.5 mol ethane = $0.5 \times 3.5 = 1.75\ \text{mol}$, which is less than the 3.5 mol supplied — so ethane is the limiting reagent and O$_2$ is in excess.
   5. Excess O$_2$ remaining = $3.5 - 1.75 = 1.75\ \text{mol}$. …

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8. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.50 mL of 1M HCl was completely reacted with x g of $\mathrm{CaCO_3}$ to form $\mathrm{CaCl_2}$, $\mathrm{CO_2}$ and $\mathrm{H_2O}$. What is the value of x in g? (Ca = 40 u; C = 12 u; O = 16 u) (A) 25 (B) 0.25 (C) 2.5 (D) 0.025

   ›Reveal solutionSolution

   Using the balanced equation's 1:2 mole ratio between CaCO$_3$ and HCl, 0.05 mol HCl reacts with 0.025 mol CaCO$_3$, i.e. 2.5 g.

   Concept and Intuition

   Stoichiometry problems require (1) the balanced equation to find mole ratios, (2) converting given volume/molarity of solution to moles, and (3) converting moles of the unknown to mass using its molar mass.

   Step-by-Step Solution

   1. Balanced equation: $\mathrm{CaCO_3(s) + 2HCl(aq) \to CaCl_2(aq) + CO_2(g) + H_2O(l)}$.
   2. Moles of HCl $= \text{Molarity}\times\text{Volume (L)} = 1\ \mathrm{M}\times0.050\ \mathrm{L} = 0.05$ mol.
   3. From the 1:2 ratio (CaCO$_3$:HCl), moles of $\mathrm{CaCO_3} = 0.05/2 = 0.025$ mol. …

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9. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.48 g of methane was completely burnt in the presence of oxygen. The liberated gas was passed into a solution containing 370 g of $\mathrm{Ca(OH)_2}$. What is the weight (in g) of $\mathrm{CaCO_3}$ formed? (Atomic weights: C = 12; H = 1; Ca = 40; O = 16) (A) 500 (B) 400 (C) 300 (D) 200

   ›Reveal solutionSolution

   Complete combustion of 48 g (3 mol) of methane gives 3 mol CO$_2$, which (with Ca(OH)$_2$ in excess) reacts 1:1 to form 3 mol = 300 g of CaCO$_3$.

   Concept and Intuition

   Complete combustion of methane follows $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, so moles of CO$_2$ produced equal moles of methane burnt. When this CO$_2$ is passed into excess Ca(OH)$_2$ (limewater), it forms the normal carbonate CaCO$_3$ in a 1:1 molar ratio (excess base prevents bicarbonate formation, which would occur only if CO$_2$ were in excess).

   Step-by-Step Solution

   1. Moles of CH$_4$ = $\dfrac{48}{16} = 3$ mol.
   2. From the balanced equation, moles of CO$_2$ produced = 3 mol.
   3. Moles of Ca(OH)$_2$ available = $\dfrac{370}{74} = 5$ mol — more than enough to consume all the CO$_2$. …

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10. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.How many litres of $O_2$ at 273 K and 1 atm pressure can give 20 L of CO in the following reaction? $C + O_2 \rightarrow CO_2 \xrightarrow{+C} CO$ (A) 40 L (B) 20 L (C) 5 L (D) 10 L

    ›Reveal solutionSolution

    The two-step reaction reduces to $2C+O_2\to2CO$, so gas volumes are in a 1:2 ratio of $O_2$ to CO; producing 20 L of CO needs only 10 L of $O_2$.

    Concept and Intuition

    When a reaction proceeds through intermediate steps, the overall stoichiometry is found by adding the steps together (cancelling species that are produced then consumed, like $CO_2$ here). Once the overall balanced equation is known, Gay-Lussac's law of combining volumes lets us directly relate gas volumes (at the same T and P) using the mole ratios from the balanced equation, without needing to compute moles explicitly.

    Step-by-Step Solution

    1. Step 1: $C + O_2 \to CO_2$.
    2. Step 2: $CO_2 + C \to 2CO$.
    3. Add the two steps (the intermediate $CO_2$ cancels): $2C + O_2 \to 2CO$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.How many grams of Mg is required to completely reduce 100 ml, 0.1 M $\mathrm{NO_3^-}$ solution using the following reaction $\mathrm{NO_3^- + Mg \rightarrow Mg^{2+} + NH_3}$ (A) 0.96 (B) 0.62 (C) 0.24 (D) 0.75

    ›Reveal solutionSolution

    A redox stoichiometry problem: balance electrons between the N reduction ($+5\to-3$) and Mg oxidation ($0\to+2$), then convert moles of Mg to grams.

    Concept and Intuition

    This is a classic redox titration-type calculation. The oxidation number of nitrogen drops from $+5$ in nitrate to $-3$ in ammonia — a gain of 8 electrons per nitrogen atom. Each magnesium atom is oxidized from $0$ to $+2$, losing 2 electrons. Electrons lost must equal electrons gained, fixing the mole ratio of Mg to NO$_3^-$ needed.

    Step-by-Step Solution

    1. Electron change per N: $+5 \to -3$ is a gain of $5-(-3)=8$ electrons.
    2. Electron change per Mg: $0 \to +2$ is a loss of 2 electrons.
    3. Electrons must balance: $2 \times (\text{mol Mg}) = 8\times(\text{mol NO}_3^-)$, so mol Mg $= 4\times$ mol NO$_3^-$.
    4. Moles of NO$_3^-$ present: $0.1\ \text{L} \times 0.1\ \text{mol/L} = 0.01$ mol. …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.A reaction gas mixture contains 50%, 30%, 20% of A, B and C by volume, respectively. The mixture undergoes the following reactions. 1. $\mathrm{A + 2B \rightarrow P_1}$ 2. $\mathrm{4P_1 + 3C \rightarrow P_2}$ $P_1$ and $P_2$ are two products of the reactions. Choose the correct answer if the reaction completes. (A) C will be completely exhausted. (B) A will be completely exhausted. (C) B will not be completely exhausted. (D) $P_1$ will be completely exhausted.

    ›Reveal solutionSolution

    Working through both reactions with a limiting-reagent analysis (basis 100 volumes: A=50, B=30, C=20) shows B is exhausted in reaction 1 and all the $P_1$ it produces is exhausted in reaction 2, while A and C both remain in excess.

    Concept and Intuition

    When a mixture undergoes sequential reactions, each step has its own limiting reagent determined by the stoichiometric ratio required versus what's actually available. The product of an earlier step (here $P_1$) can itself become the limiting reagent of a later step, even if none of the original reactants directly limit that later step.

    Step-by-Step Solution

    1. Take a convenient basis of 100 volume units: A = 50, B = 30, C = 20.
    2. Reaction 1: $A + 2B \to P_1$ requires a 1:2 mole ratio of A:B. To consume all of B (30), only $30/2 = 15$ mol of A is needed. Since 50 mol A is available (> 15), B is limiting.
    3. B is completely consumed; A remaining $= 50 - 15 = 35$ mol (in excess); $P_1$ produced $= 15$ mol (equal to A consumed, by 1:1 stoichiometry with A).
    4. Reaction 2: $4P_1 + 3C \to P_2$ requires a 4:3 mole ratio of $P_1$:C. To consume all 15 mol $P_1$, only $15 \times \frac34 = 11.25$ mol C is needed. Since 20 mol C is available (> 11.25), $P_1$ is limiting.
    5. $P_1$ is completely consumed; C remaining $= 20 - 11.25 = 8.75$ mol (in excess). …

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