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?

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%