Question

In an experiment, $6.67$ g of $AlC{l}_3$ was produced and $0.54$ g of $Al$ remained unreacted. How many gram atoms of $Al$ and $C{l}_2$ were taken originally? (Given that atomic weight of $Al$ is $27$ g and of $Cl$ is $35.5$ g.)

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

Answer 1

The correct option is A $0.07,0.15$

Moles of $AlC{l}_3$ produced $=\frac{6.67}{133.5}=0.05$ mol
Excess of $Al=\frac{0.54}{27}=0.02$ moles
Gram atoms or moles of $Al$ taken $=0.05+0.02=0.07$
Gram atoms or moles of $C{l}_2$ taken $=3\times 0.05=0.15$