Question

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

$\underset{2mol}{2Al}+6HCl⟶\underset{3mol}{2AlC{l}_3+3H_2(g)}$

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

$=3.3825c{m}^3$

Density $=2.70g/c{m}^3$

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

$=9.13275g$

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

$=0.32825mol$

$H_2$ formed $=1.5$ times of $Al$ reacted

$=1.5\times 0.33825=0.51mol$