Question

Calcium carbonate reacts with aqueous HCl to give $CaC{l}_2$ and $C{O}_2$ according to the reaction:
$CaC{O}_3\left(s\right)+HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$
When 250 mL of 0.76 M HCl reacts with 1000 g of $CaC{O}_3$, calculate the number of moles of $CaC{l}_2$ formed:

• A. $0.19mol$
• B. $0.095mol$
• C. $1.20mol$
• D. $2.41mol$

Answer 1

The correct option is B $0.095mol$

Balanced chemical equation
$CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$
Moles of $CaC{O}_3$ = $\frac{Mass}{Molarmass}$
= $\frac{1000g}{100g/mol}$
= $10mol$
Moles of $HCl$ = Molarity $\times$ Volume of solution (L)
= $0.76M\times 0.25L$
$=0.19mol$
For calculation of limiting reagent = $\frac{\text{Moles of reactant}}{\text{Stoichiometry of the reactant}}$
For $CaC{O}_3$ the value is $\frac{10}{1}$ = 10
For $HCl$ the value is $\frac{0.19}{2}$ = 0.095
Hence $HCl$ is limiting reagent.
According to balanced chemical equation, 2 mol of $HCl$ produce 1 mol of $CaC{l}_2$
By unitary method, 0.19 mol of $HCl$ will produce $\frac{1}{2}\times 0.19$ = 0.095 mol