Question

In an experiment, $6.67\, g$ of $AlCl _3$ was produced and $0.54\, g\, Al$ remained unreacted. How many $g$ atoms of $Al$ and $Cl _2$ were taken originally $( Al =27, Cl =35.5)$ ?

• A. $0.07,0.15$
• B. $0.07,0.05$
• C. $0.02,0.05$
• D. $0.02,0.15$

Answer 1

Answer: (A)

Moles of $AlCl _3$ produced $=\frac{6.67}{133.5}=0.05\, mol$
Excess of $Al =\frac{0.54}{27}=0.02$ mole
$g$ atom or moles of $Al$ taken $=0.05+0.02=0.07$
$g$ atom or moles of $Cl _2$ taken $=3 \times 0.05=0.15$