Question

A $15.0g$ sample of lithium is reacted with $15.0g$ of fluorine to form lithium fluoride:
$2Li+F_2\to 2LiF$
After the reaction is complete, what will be present?

• A. $2.16molesLiF$ only
• B. $0.789molesLiF$ only
• C. $2.16molesLiF$ and $0.395moles{F}_2$
• D. $0.789moles$ of $LiF$ and $1.37molesLi$

Answer 1

The correct option is D $0.789moles$ of $LiF$ and $1.37molesLi$

$2Li+F_2⟶2LiF$

$15g$ $15g$

Number of moles of $F_2=\frac{15}{38}=0.395,$

Moles of $Li$ required $=0.395\times 2=\mathrm{0.790.}$

Moles of $Li=\frac{15}{6.9}=2.17$

$Li$ is in excess and $F_2$ is the limiting reagent.

Moles of $Li$ in excess $=2.17-0.790=1.37$

Moles of $LiF=0.395\times 2=0.790$

So, after completion, $0.79$ moles of $LiF$ and $1.37$ moles of $Li$ will be present.