Question

A piece of aluminium foil measuring $10.25 \,cm \times 5.50 \,cm$ $\times 0.600\, mm$ is dissolved in excess of $HCl$ (aq). Density of foil is $2.70\, g /\, cm ^{3}$. Thus, gaseous product formed is

• A. 0.51 mol
• B. 1.52 mol
• C. 1.01 mol
• D. 3.00 mol

Answer 1

Answer: (A)

$\underset{2\, mol}{2 Al} +6 HCl \longrightarrow \underset{3\,mol}{2 AlCl _{3}+3 H _{2}( g )}$

Volume of foil $=10.25\, cm \times 5.50 \,cm \times \frac{0.600}{10} \,cm$

$=3.3825 \,cm ^{3}$

Density $=2.70 \,g / cm ^{3}$

$\therefore $ Mass of $ Al =3.3825 \times 2.70$

$=9.13275\, g $

Moles of $Al =\frac{9.13275}{27}$

$=0.32825\,mol$

$H _{2}$ formed $=1.5$ times of $Al$ reacted

$=1.5 \times 0.33825=0.51\, mol$