Question

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actually amount of CuS produced was 3.05 g. Reaction: Na2S + CuSO4 → Na2SO4 + CuS (A) 16.1% (B) 42.1% (C) 18.93% (D) 7.25% (E) not enough information

Answer 1

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Na_2S$

Given mass = 15.5 g

Molar mass of $Na_2S$ = 78.0452 g/mol

Moles of $Na_2S$ = 15.5 g / 78.0452 g/mol = 0.1986 moles

Given: For $CuSO_4$

Given mass = 12.1 g

Molar mass of $CuSO_4$ = 159.609 g/mol

Moles of $CuSO_4$ = 12.1 g / 159.609 g/mol = 0.0758 moles

According to the given reaction:

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

1 mole of $Na_2S$ react with 1 mole of $CuSO_4$

0.1986 mole of $Na_2S$ react with 0.1986 mole of $CuSO_4$

Available moles of $CuSO_4$ = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, $CuSO_4$ is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuSO_4$ gives 1 mole of $CuS$

0.0758 mole of $CuSO_4$ gives 0.0758 mole of $CuS$

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

Option B is correct.