Answer:
% yield of reaction is 26.4
Explanation:
The reaction is:
Mg + I₂ → MgI₂
Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:
10 g . 1mol / 24.3 g = 0.411 moles of Mg
95 g . 1mol / 253.8g = 0.374 moles of I₂
Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine
For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.
As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂
We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g
To calculate the percent yield:
% yield = (yield produced /theoretical yield) . 100
% yield = (27.5 g/ 104g) . 100 = 26.4
