In dilute sulfuric acid was dissolved 12.35 grams of zinc, what is the volume of hydrogen if the percentage of output is 85%.

1. We compose the reaction equation:

Zn + H2SO4 = ZnSO4 + H2

According to the equation: 1 mole of zinc entered the reaction. Let’s find its mass by the formula: m = n * M

m (Zn) = 1 mol * 65 g / mol = 65 g

According to the equation: 1 mole of hydrogen was released, i.e. 22.4 l

2. We make the proportion:

12.35 g zinc – x l hydrogen

65 g zinc – 22.4 l hydrogen

Hence, x = 12.35 * 22.4 / 65 = 4.256 liters.

3. Let us calculate the practical output of H2:

4.256 L – theoretical yield – 100%

xl – practical yield – 85%

Hence, x = 4.256 * 85/100 = 3.62 liters of H2.