Answer:
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 %
