Question

How many grams of sodium carbonate is obtained on heating $42g$ of $NaHC{O}_{3\left(s\right)}$?

• A. $21g$
• B. $26.5g$
• C. $42g$
• D. $46g$

Answer 1

The correct option is B $26.5g$

$2NaHC{O}_3\stackrel{\text{Heat}}{\to }N{a}_{2}C{O}_3+C{O}_2+H_{2}O$
Given mass of $NaHC{O}_3$ = $42g$
Molar mass of $NaHC{O}_3$ = $84g$
Number of moles of $NaHC{O}_3$ = $\frac{42}{84}$
= $0.5\text{moles}$
$2\text{moles}$ 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 $0.5\text{moles}$ of $NaHC{O}_3$ = $\frac{0.5}{2}=0.25moles$

Mass of $N{a}_{2}C{O}_3$ produced
$=\text{Number of moles of}N{a}_{2}C{O}_3\text{produced}\times \text{Molar mass of}N{a}_{2}C{O}_3$
= $0.25\text{moles}\times 106{\text{g mol}}^{-1}$
= $26.5\text{g}$