Question

The mass of h2 produced by reaction of 1.80 g al and 6.00 g h2so4 is 0.112 g. what is the percent yield?

Answer 1

First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:

2 Al + 3 H₂SO₄ = 3 H₂ + Al₂(SO₄)₃

Next, let us determine which of the reactants is limiting:

1.80 g Al (1 mol/26.98 g)(3 mol H₂SO₄/2 mol Al)(98 g H₂SO₄/mol) = 9.807 g

It would need 9.807 g. But the only available amount is 6 g. That means that H₂SO₄ is the limiting reactant. We use 6 g as the basis to know the theoretical yield:

6 g*(1 mol H₂SO₄/98 g)*(3 mol H₂/ 3 mol H₂SO₄)*(2 g /mol H₂) = 0.122 g

Therefore, the percent yield is equal to
Percent yield = (0.112 g/0.122 g)*100
Percent yield = 91.8%