Chemistry - Stoichiometry Question with Solution | TestHub

ChemistryStoichiometryProblems Based on Mixtures, Percentage PurityMedium2 minQB
ChemistryMediummatching list

Match the entries in Column I with entries in Column II and then pick out correct options.

Column I Column II

(P) (1) Moles of gas = 2.4

(Q) (2) Volume of gas at 1 atm, 273K

(R) (3) Moles of reactant (left) = 1.8 mol

(S) (4) Moles of solid product = 1.8 mol

 

Options:

Answer:
C
Solution:

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:

Initial moles of A = 3 mol

Percentage yield = 40%

 

Moles of A reacted =

 

From the stoichiometry of the reaction:

2 mol A produces 4 mol C(g)

So, 1.2 mol A will produce

 

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

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

Volume of C(g) = . This matches.

(3) Moles of reactant (left) = Initial moles of A - Moles of A reacted

Moles of A left = . 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 . This matches.

 

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

 

Part Q:

Reaction:

Initial moles of A = 2 mol

Percentage yield = 60%

 

Moles of A reacted =

 

From the stoichiometry of the reaction:

1 mol A produces 2 mol B(g)

So, 1.2 mol A will produce

 

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

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

Volume of B(g) = . This matches.

(3) Moles of reactant (left) = Initial moles of A - Moles of A reacted

Moles of A left = . 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:

Initial moles of B = 1.2 mol

Percentage yield = 50%

 

Moles of B reacted =

 

From the stoichiometry of the reaction:

1 mol B produces 4 mol C(g)

So, 0.6 mol B will produce

 

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

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

Volume of C(g) = . This matches.

(3) Moles of reactant (left) = Initial moles of B - Moles of B reacted

Moles of B left = . 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 . This does not match 1.8 mol.

 

Thus, R matches with 1, 2.

 

Part S:

Reaction:

Initial moles of C = 1.5 mol

Percentage yield = 80%

 

Moles of C reacted =

 

From the stoichiometry of the reaction:

1 mol C produces 2 mol E(g)

So, 1.2 mol C will produce

 

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

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

Volume of E(g) = . This matches.

(3) Moles of reactant (left) = Initial moles of C - Moles of C reacted

Moles of C left = . 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 . This does not match 1.8 mol.

 

Thus, S matches with 1, 2.

 

Summary of matches:

P 1, 2, 3, 4

Q 1, 2

R 1, 2

S 1, 2

 

Comparing this with the given options:

A. P1,4; Q2,3; R2,3; S1,2,3 (Incorrect)

B. P3,4; Q3,4; R2,4; S1,2 (Incorrect)

C. P1,2,3,4; Q1,2; R1,2; S1,2 (Correct)

D. P1,2,4; Q1,2,3; R2; S1,2 (Incorrect)

 

The final answer is .

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Problems Based on Mixtures, Percentage Purity
2mℹ️ Source: QB

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