Question

Some commercial drain cleaners contain two components: sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: 2NaOH(aq) + 2Al(s) + 6H2O(l)  2NaAl(OH)4 (aq) + 3H2(g) The heat generated in this reaction helps to melt away grease and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of H2 formed at 298 K and 1.1 atm if 1.34 g of Al is treated with excess NaOH and H2O

Answer 1

The balanced equation is-

2NaOH (aq) + 2Al(s) + 6H2O(l) = 2NaAl(OH)4 (aq) + 3H2(g)

So, 3 moles hydrogen is produced from 2 moles of Aluminium.

Molar Mass of Al = 27 g/mol. Therefore, 1.34 g Al = 1.34/27 mol

Moles of H2 gas produced = (3/2)*(1.34/27) = 0.0744

From the combined gas laws we know that,

PV = nRT

Or, V = nRT/P, where, P = 1.1 atm, T= 298 K, R = 0.082 L atm / mol K, n = 0.0744 mol

By substituting the values, we get,

V = 0.0744*0.082*298/1.1 L = 1.653 L

Volume of H₂ formed = 1.653 L