What volume of hydrogen can be obtained by the action of 100 g of 4.9% sulfuric acid solution per 10 g

What volume of hydrogen can be obtained by the action of 100 g of 4.9% sulfuric acid solution per 10 g of technical zinc containing 20% of impurities, if the volume fraction of hydrogen yield is 80% of the theoretically possible?

The interaction of zinc with sulfuric acid occurs in accordance with the following chemical reaction equation:

Zn + H2SO4 = ZnSO4 + H2;

When 1 mole of metal is dissolved, 1 mole of hydrogen gas is released.

Let’s calculate the amount of a substance contained in a weighed portion of 10 grams of technical zinc containing 20% ​​impurity.

To do this, we will find the ratio of its weight to molar mass.

m Zn = 10 x 0.8 = 8 grams;

M Zn = 65 grams / mol;

N Zn = 8/65 = 0.0123 mol;

Find the amount of substance in 100 grams of 4.9% sulfuric acid solution

Its mass is equal to:

m H2SO4 = 100 x 0.049 = 4.9 grams;

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 4.9 / 98 = 0.05 mol;

0.0123 mol of zinc will react with 0.0123 mol of sulfuric acid. This will release 0.0123 mol of hydrogen. Taking into account the fact that the volume fraction of hydrogen yield is 80% of the theoretically possible, 0.0123 x 0.8 = 0.00984 mol of hydrogen will be released.

Let’s calculate its volume.

One mole of gas fills a volume of 22.4 liters under normal conditions.

The volume of hydrogen will be:

V H2 = 0.00984 x 22.4 = 0.22 liters;