A student completes a lab and produces 1.3 moles of hydrogen gas by adding magnesium to sulfuric acid. If the student started with 22.8 grams of magnesium, what was the student's percent yield?

72 %

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let's gather all the information in one place.

M_r: 24.30

Mg + H₂SO₄ ⟶ MgSO₄ + H₂

m/g: 22.8

n/mol: 1.3

Calculations:

(a) Moles of Mg

n = 22.8 g Mg * (1 mol Mg/24.30 g Mg)

= 0.9383 mol Mg

(b) Moles of H₂

The molar ratio is (1 mol H₂/1 mol Mg).

n = 0.9383 mol Mg * (1 mol H₂/1 mol Mg)

= 0.9383 mol H₂

(c) Percent yield

% yield = actual yield/theoretical yield * 100 %

= 0.9383 mol/1.3 mol * 100 %

= 72 %