Calculate the volume of hydrogen and the mass of salt that will be released during the interaction

Calculate the volume of hydrogen and the mass of salt that will be released during the interaction of magnesium with sulfuric acid weighing 200 grams with a mass fraction of 40%

Magnesium interacts with sulfuric acid. In this case, magnesium sulfate is synthesized and hydrogen gas is released. This reaction is described by the following chemical reaction equation:

Mg + H2SO4 = MgSO4 + H2;

In accordance with the coefficients of the equation, when 1 mol of magnesium is dissolved, 1 mol of gaseous hydrogen is released.

Let’s calculate the chemical amount of sulfuric acid. To do this, divide the available weight of the acid by the weight of 1 mole of acid.

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

N H2SO4 = 200 x 0.4 / 98 = 0.816 mol;

The same amount of hydrogen will be synthesized.

Let’s calculate its volume, for this we multiply the amount of substance by the volume of 1 mole of gas (22.4 liters):

V H2 = 0.816 x 22.4 = 18.28 liters;

We calculate the weight of 0.816 mol of magnesium sulfate.

M MgSO4 = 24 + 32 + 16 x 4 = 120 grams / mol;

m MgSO4 = 0.816 x 120 = 97.92 grams;