Question

If 20.0 g of $CaC{O}_3$ is treated with 20.0 g of $HCl$, how many grams of $C{O}_2$ can be produced according to the reaction

$CaC{O}_3\left(s\right)+2HCl\left(aq\right)$

$\downarrow$

$CaC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

• A. 8.00 g
• B. 7.70 g
• C. 7.20 g
• D. 8.80 g

Answer 1

The correct option is D 8.80 g

Identifying the limiting reagent:

Moles of $CaC{O}_3=\frac{mass}{molarmass}=\frac{20}{100}=0.2mol$

$MolesofHCl=\frac{mass}{molarmass}=\frac{20}{36.5}=0.55mol$

Chemical reaction involved:
$CaC{O}_3\left(s\right)+2HCl\left(aq\right)$

$\to CaC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

According to the reaction,
1 mol of $CaC{O}_3$ reacts with 2 mol of $HCl$.
Thus, 0.2 mol of $CaC{O}_3$ will react with 2 × 0.2
i.e., 0.4 mol of $HCl$.

The given moles of $HCl$ is 0.55. This implies that $HCl$ is present in excess and $CaC{O}_3$ is the limiting reagent. Therefore, $CaC{O}_3$ will control the amount of product formed.

Calculating the amount of $C{O}_2$ obtained:

$CaC{O}_3\left(s\right)+2HCl\left(aq\right)$

$\downarrow$

$CaC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

According to the reaction given,
1 mol of $CaC{O}_3$ produces 1 mol of $C{O}_2$.
Thus, 0.2 mol of $CaC{O}_3$ will produce 0.2 mol of $C{O}_2$.

Mass of $C{O}_2$ = number of moles × molar mass
= 0.2 mol × $44gmo{l}^{-1}$
= 8.8 g
Hence, the amount of $C{O}_2$ produced is 8.8 g.
So, option (a) is the correct answer.