Astudent 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 %

the answer is the first option

explanation:

just googled it m8. not that hard

lol