Question

The weight of $NaHC{O}_3$, that can give the same weight of $C{O}_2$, which can be prepared from 12.5 g of limestone is:

• A. $21g$
• B. $10g$
• C. $84g$
• D. $11g$

Answer 1

The correct option is A $21g$

Decomposition of $CaC{O}_3$ can be written as:
$CaC{O}_3\to CaO+C{O}_2$

12.5g of $CaC{O}_3$ contains = $\frac{12.5}{100}$ = 0.125 moles of $CaC{O}_3$

From the above equation we know;
$0.125$ moles of $CaC{O}_3$ $=0.125$ moles of $C{O}_2$.

Now the equation for the decomposition of $NaHC{O}_3$ can be written as:
$2NaHC{O}_3\to N{a}_{2}C{O}_3+H_{2}O+C{O}_2$

$\therefore$ 1 mole of $C{O}_2$ is formed from = 2 moles of $NaHC{O}_3$
0.125 moles of $C{O}_2$ are formed from = $2\times 0.125$ $=0.25$ moles of $NaHC{O}_3$

0.25 moles of $NaHC{O}_3$ = $0.25\times 84$ $=21$ g of $NaHC{O}_3$

Hence, option A is correct.