Question

42. What is the percent yield if 68.9 g of silver metal were recovered from a solution containing 105.7 g of Agt reacts with 35.2 g of Mg? Mg(s) + 2 Ag+ (aq) → Mg2+ (aq) + 2 Ag(s)

Answer 1

Mg + 2 Ag+ -----------------> Mg2 + + 2 Ag

no of moles = mass /molar mass

for Mg = 35.2 g / 24.305 g/mol => 1.448 mol

for Ag+ = 105.7 g / 107.8682 g/mol => 0.9798 mol

the mole ratio is 1 : 1

so lesser moles having the reactant acts as limiting reactant

limiting reactant is Ag+

theoretical yield = 0.9798 mol * 170.8682 g/mol => 105.7 grams

percent yield = actual yield / theoretical yield * 100

percent yield = 68.9 / 105.7 *100 => 65.2 %