Question

If $21\text{g}$ of $NaHC{O}_3$ is added to $15\text{g}$ of $C{H}_{3}COOH$ solution, the residue is found to weigh $25\text{g}$. what is the mass of $C{O}_2$ is released in the reaction?

• A. $11\text{g}$
• B. $12\text{g}$
• C. $13\text{g}$
• D. $10\text{g}$

Answer 1

The correct option is A $11\text{g}$

Before the reaction:
$\underset{21\text{g}}{NaHC{O}_3}+\underset{15\text{g}}{C{H}_{3}COOH}\to C{H}_{3}COONa+H_{2}O+C{O}_2$
Total mass before the reaction $=21+15$
Mass of the residue after the reaction$=25g$
let $x$ gram of $C{O}_2$ be liberated
Total mass after reaction $=25+x$
According to the law of conservation of mass,
$\Delta m=0$
i.e., $massafterthereaction-massbeforethereaction=0\left(i\right)$
$25+x-36=0$
or $x=36-25=11\text{g}$
Therefore, mass of $C{O}_2$ released = $11\text{g}$