Chemistry - Stoichiometry Question with Solution | TestHub

Explanation:

This problem involves stoichiometry and the calculation of moles and volumes of gases based on given reactions and their percentage yields. We will analyze each reaction (P, Q, R, S) and match them with the corresponding statements (1, 2, 3, 4).

Given: Molar volume of gas at 1 atm and 273 K (STP) is 22.4 L/mol.

Part P:

Reaction: $2\mathrm{A}(\mathrm{s})\stackrel{40\%}{\to }3\mathrm{B}(\mathrm{s})+4\mathrm{C}(\mathrm{g})$Initial moles of A = 3 mol Percentage yield = 40%

Moles of A reacted = $3mol\times 0.40=1.2mol$

From the stoichiometry of the reaction: 2 mol A produces 4 mol C(g)

So, 1.2 mol A will produce $\frac{4}{2}\times 1.2molC(g)=2\times 1.2molC(g)=2.4molC(g)$

(1) Moles of gas = 2.4 mol. This matches.

(2) Volume of gas at 1 atm, 273 K = Moles of gas $\times$ Molar volume

Volume of C(g) = $2.4mol\times 22.4L/mol=53.76L$. This matches. (3) Moles of reactant (left) = Initial moles of A - Moles of A reacted

Moles of A left = $3mol-1.2mol=1.8mol$. This matches. (4) Moles of solid product = Moles of B(s) formed From stoichiometry: 2 mol A produces 3 mol B(s)

So, 1.2 mol A will produce $\frac{3}{2}\times 1.2molB(s)=1.8molB(s)$. This matches.

Thus, P matches with 1, 2, 3, 4.

Part Q:

Reaction: $\mathrm{A}(\mathrm{s})\stackrel{60\%}{\to }2\mathrm{B}(\mathrm{g})$Initial moles of A = 2 mol Percentage yield = 60%

Moles of A reacted = $2mol\times 0.60=1.2mol$

From the stoichiometry of the reaction: 1 mol A produces 2 mol B(g)

So, 1.2 mol A will produce $2\times 1.2molB(g)=2.4molB(g)$

(1) Moles of gas = 2.4 mol. This matches.

(2) Volume of gas at 1 atm, 273 K = Moles of gas $\times$ Molar volume

Volume of B(g) = $2.4mol\times 22.4L/mol=53.76L$. This matches. (3) Moles of reactant (left) = Initial moles of A - Moles of A reacted

Moles of A left = $2mol-1.2mol=0.8mol$. This does not match 1.8 mol. (4) Moles of solid product = 0 (since B is a gas). This does not match 1.8 mol.

Thus, Q matches with 1, 2.

Part R:

Reaction: $\mathrm{B}(\mathrm{s})\stackrel{50\%}{\to }4\mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{s})$Initial moles of B = 1.2 mol Percentage yield = 50%

Moles of B reacted = $1.2mol\times 0.50=0.6mol$

From the stoichiometry of the reaction: 1 mol B produces 4 mol C(g)

So, 0.6 mol B will produce $4\times 0.6molC(g)=2.4molC(g)$

(1) Moles of gas = 2.4 mol. This matches.

(2) Volume of gas at 1 atm, 273 K = Moles of gas $\times$ Molar volume

Volume of C(g) = $2.4mol\times 22.4L/mol=53.76L$. This matches. (3) Moles of reactant (left) = Initial moles of B - Moles of B reacted

Moles of B left = $1.2mol-0.6mol=0.6mol$. This does not match 1.8 mol. (4) Moles of solid product = Moles of D(s) formed From stoichiometry: 1 mol B produces 1 mol D(s)

So, 0.6 mol B will produce $1\times 0.6molD(s)=0.6molD(s)$. This does not match 1.8 mol.

Thus, R matches with 1, 2.

Part S:

Reaction: $\mathrm{C}(\mathrm{s})\stackrel{80\%}{\to }\mathrm{D}(\mathrm{s})+2\mathrm{E}(\mathrm{g})$Initial moles of C = 1.5 mol Percentage yield = 80%

Moles of C reacted = $1.5mol\times 0.80=1.2mol$

From the stoichiometry of the reaction: 1 mol C produces 2 mol E(g)

So, 1.2 mol C will produce $2\times 1.2molE(g)=2.4molE(g)$

(1) Moles of gas = 2.4 mol. This matches.

(2) Volume of gas at 1 atm, 273 K = Moles of gas $\times$ Molar volume

Volume of E(g) = $2.4mol\times 22.4L/mol=53.76L$. This matches. (3) Moles of reactant (left) = Initial moles of C - Moles of C reacted

Moles of C left = $1.5mol-1.2mol=0.3mol$. This does not match 1.8 mol. (4) Moles of solid product = Moles of D(s) formed From stoichiometry: 1 mol C produces 1 mol D(s)

So, 1.2 mol C will produce $1\times 1.2molD(s)=1.2molD(s)$. This does not match 1.8 mol.

Thus, S matches with 1, 2.

Summary of matches:

P $\to$ 1, 2, 3, 4

Q $\to$ 1, 2

R $\to$ 1, 2

S $\to$ 1, 2

Comparing this with the given options:

A. P$\to$1,4; Q$\to$2,3; R$\to$2,3; S$\to$1,2,3 (Incorrect)

B. P$\to$3,4; Q$\to$3,4; R$\to$2,4; S$\to$1,2 (Incorrect)

C. P$\to$1,2,3,4; Q$\to$1,2; R$\to$1,2; S$\to$1,2 (Correct)

D. P$\to$1,2,4; Q$\to$1,2,3; R$\to$2; S$\to$1,2 (Incorrect)

The final answer is $\boxed{C}$.