Question

How many grams of sodium carbonate is obtained by heating $8.4g$ of $NaHC{O}_3\left(s\right)$?

• A. $2.1g$
• B. $5.3g$
• C. $4.2g$
• D. $4.6g$

Answer 1

The correct option is B $5.3g$

$2NaHC{O}_3\stackrel{\text{Heat}}{\to }N{a}_{2}C{O}_3+C{O}_2+H_{2}O$
Given mass of $NaHC{O}_3$ = $8.4g$
Molar mass of $NaHC{O}_3$ = $84g$
Number of moles of $NaHC{O}_3$ = $\frac{8.4}{84}$
= $0.1\text{moles}$
$2\text{moles}$ of $NaHC{O}_3$ gives $1\text{mole}$ of $N{a}_{2}C{O}_3$ on heating.

Number of moles of $N{a}_{2}C{O}_3$ produced on heating $0.1\text{moles}$ of $NaHC{O}_3$ = $\frac{0.1}{2}=0.05moles$

Mass of $N{a}_{2}C{O}_3$ produced
$=\text{Number of moles of}N{a}_{2}C{O}_3\text{produced}\times \text{Molar mass of}N{a}_{2}C{O}_3$
= $0.05\text{moles}\times 106{\text{g mol}}^{-1}$
= $5.3\text{g}$