Chemistry - Stoichiometry Question with Solution | TestHub

The reactions are:

$$2\mathrm{Al}+6\mathrm{HCl}\to 2{\mathrm{AlCl}}_3+3{\mathrm{H}}_2$$

$$\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$$

Step 1: Calculate moles of reactants for the first reaction.

Moles of Al = $5.4\mathrm{g}/27g/mol=0.2\mathrm{mol}$

Moles of HCl = $0.200\mathrm{L}\times 3.0mol/L=0.6\mathrm{mol}$

From the stoichiometry $2\mathrm{Al}:6\mathrm{HCl}$, $0.2\mathrm{mol}$ of Al requires $0.2\times (6/2)=0.6\mathrm{mol}$ of HCl. Both reactants are completely consumed.

Moles of H₂ produced = $0.2\mathrm{mol}\times (3/2)=0.3\mathrm{mol}$.

Step 2: Calculate moles of reactants for the second reaction.

Moles of CuO = $15.9\mathrm{g}/79.5g/mol=0.2\mathrm{mol}$

Moles of H₂ available from Step 1 = $0.3\mathrm{mol}$.

From the stoichiometry $1\mathrm{CuO}:1{\mathrm{H}}_2$, $0.2\mathrm{mol}$ of CuO requires $0.2\mathrm{mol}$ of H₂.

Since $0.3\mathrm{mol}$ H₂ is available, CuO is the limiting reagent.

Moles of Cu produced = $0.2\mathrm{mol}$ (from $0.2\mathrm{mol}$ CuO).

Mass of Cu = $0.2\mathrm{mol}\times 63.5g/mol=12.7\mathrm{g}$.