0.81 g of aluminum was added to a solution containing 4.9 g of sulfuric acid. How much hydrogen

0.81 g of aluminum was added to a solution containing 4.9 g of sulfuric acid. How much hydrogen will be released during the reactions?

Aluminum, as an active metal, displaces hydrogen from the acid.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2.

Let us calculate the number of moles of the starting materials taking into account the stoichiometric coefficients of the reaction.

n (Al) = m / Mr • 2 = 0.81 g / 27 g / mol • 2 = 0.015 mol.

n (H2SO4) = m / Mr • 3 = 4.9 g / 98 g / mol • 3 = 0.016 mol.

Sulfuric acid is in excess, so the calculation is based on aluminum.

The number of moles of hydrogen is.

n (H2) = n (Al) • 3 = 0.015 mol • 3 = 0.045 mol.

This means that the volume of released hydrogen is equal to.

V (H2) = n • Vm = 0.045 mol • 22.4 mol / L = 1.008 L.