Moles, Chemistry

Question

3.3 grams of MgO is produced by the reaction between 3.0 grams of Mg and 15 liters of O₂ . Determine the percent yield of MgO .

Answer 1

Hi Brian,

You need to have a balanced chemical equation to do any sort of stoichiometry problem.

The reaction is 2 Mg + O₂ --> 2 MgO

To find the percent yield you need to know which reactant is limiting the equation. (take BOTH and solve for grams of MgO)

15 L O₂ x 1 mol O₂ x 2 mol MgO x 40.30 g MgO = 53.97 g MgO

22.4 L O₂ 1 mol O₂ 1 mol MgO

3.0 g Mg x 1 mol Mg x 2 mol MgO x 40.30 g MgO = 4.97 g MgO

24.31 g Mg 2 mol Mg 1 mol MgO

Since the magnesium is limiting the equation...use that theoretical yield to find the percent yield

% yield = Actual yield x 100% % yield = 3.3 g MgO x 100% = 66.4% yield (66% to 2 s.f.)

Theoretical yield 4.97 g MgO