Calculate the mass of aluminum that will need to be dissolved in an excess of sodium hydroxide

Calculate the mass of aluminum that will need to be dissolved in an excess of sodium hydroxide solution to produce 4.48L of hydrogen.

1. Let’s compose the equality of chemical interaction:

2NaOH + 2Al + 2H2O = 2Na [Al (OH) 4] + H2.

2. Let’s calculate the number of moles of the formed hydrogen (Vm – molar volume, constant equal to 22.4 l / mol):

n (H2) = V (H2) / Vm = 4.48 L / 22.4 L / mol = 0.2 mol.

3. Using the equality of the chemical interaction, we calculate the number of moles and, as a result, the mass of the reacted aluminum:

n (Al) = n (H2) * 2 = 0.2 mol * 2 0.4 mol.

m (Al) = n (Al) * M (Al) = 0.4 mol * 27 g / mol = 10.8 g.

Result: m (Al) = 10.8 g.