Question

(i) $4Li+O_2\to 2L{i}_{2}O$
$21.0$ g of lithium reacts with $32.0$ g of $O_2$.

(ii) $2K+C{l}_2\to 2KCl$

$3.9$ g of $K$ reacts with $4.26$ g of $C{l}_2$.
[The atomic weight of $Li=7$ g and $K=39$ g while molecular weight of $L{i}_{2}O=30$ g/mol and $KCl=74.5$ g/mol.]

Which of the following statements is/are correct?

• A. $O_2$ is in excess in reaction (i).
• B. $45.0$ g of $L{i}_{2}O$ is formed is reaction (i).
• C. $C{l}_2$ is in excess in reaction (ii).
• D. $7.45$ g of $KCl$ is formed is reaction (ii).

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $45.0$ g of $L{i}_{2}O$ is formed is reaction (i).
B $C{l}_2$ is in excess in reaction (ii).
C $O_2$ is in excess in reaction (i).
D $7.45$ g of $KCl$ is formed is reaction (ii).

Moles of $Li=\frac{21}{7}=3.0$ mol and moles of $O_2=\frac{32}{32}=3.0$ mol

A. Since, $\frac{3}{4}=0.75$ mol of $O_2$ is required, therefore, $(1.0-0.75)=0.25$ mol of $O_2$ is in excess. Hence, $Li$ is the limiting reagent.

B. Weight of $L{i}_{2}O$ formed $=\frac{3\times 1\times 30}{2}=45.0$ g $L{i}_{2}O$

C. Moles of $K=\frac{3.9}{39}=0.1$ mol and moles of $C{l}_2=\frac{4.26}{71}=0.06$ mol
Since, $\frac{0.1}{2}=0.05$ mol of $C{l}_2$ is required. Therefore, $(0.06-0.05)=0.01$ mol of $C{l}_2$ is in excess.
Hence, $K$ is the limiting reagent.

D. Weight of $KCl$ formed $=0.1\times 1\times 74.5=74.5$ g of $KCl$

Hence, options A, B, C and D are correct.