What volume of hydrogen can be obtained by the interaction of magnesium weighing 1.2 g

What volume of hydrogen can be obtained by the interaction of magnesium weighing 1.2 g with a solution containing 5.5 g of sulfuric acid?

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, upon dissolution of 1 mol of magnesium, 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 = 5.5 / 98 = 0.056 mol;

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

M Mg = 24 grams / mol;

N Mg = 1.2 / 24 = 0.05 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 = 0.05 x 22.4 = 1.12 liters;