Question

2. What is the percent yield if 13.5 g SO2 are obtained from the reaction of 21.3 g of O2 with 28.6g of ZnS according to the following equation? 2 ZnS (s) + 3 O2 (g) → 2 ZnO (s) + 2 SO2 (g) 3. Calculate the percent yield of KCl if 1.043 g of KCI are produced from the reaction of 2.112 g of K2COs with 135 mL of 0.255 M HCl according to the following reaction. K2CO3 + 2 HCl → 2 KCl + CO2 H2O 32

Can you help me solve these problems please and post how to set up problem to get to thr answers? Please and Thank You.

Answer 1

2. The percent yield of SO₂ is 72.7%.

From the given equation, 2 ZnS + 3 O₂ → 2 ZnO + 2 SO₂ the given amount of ZnS, O₂ and SO₂ is 28.6 g, 21.3 g and 13.5 g respectively.

Molecular weight of ZnS and O₂ are 97.47 gmol^-1 and 32 gmol^-1.

Therefore, we have 28.6 g/ 97.47 gmol^-1 = 0.29 mol of ZnS, and

                                     21.3 g/ 32 gmol^-1 = 0.67 mol of O_2.

it is evident that 2 moles of ZnS reacts with 3 mole of O₂ to give 2 mole of SO₂

therefore, 0.29 mol of ZnS will react with (3 x 0.29) / 2 = 0.44 mol of O₂ to give 0.29 mol of SO₂

Molecular weight of SO₂ is 64 gmol^-1 .

Hence, theoretically 28.6 g of ZnS or 0.29 mol of ZnS should give O.29 mol = 0.29 mol x 64 gmol^-1 = 18.56 g of SO₂.

Finally, % Yield = (actual yield/theoretical yield) x 100

                            = (13.5 g/ 18.56 g) x 100 = 72.7 %