Question

$50g$ of $CaC{O}_3$ is allowed to react with $70g$ of $H_{3}P{O}_4$. Calculate
(i) amount of $C{a}_3(P{O}_4{)}_2$ formed and (ii) amount of unreacted reagent.

• A. (i) $51.7g$ , (ii) $40.1g$
• B. (i) $23.1g$ , (ii) $87.3g$
• C. (i) $51.7g$ , (ii) $37.3g$
• D. $Noneoftheabove$

Answer 1

The correct option is C (i) $51.7g$ , (ii) $37.3g$

The balanced chemical equation is as follows:
$3CaC{O}_3+2H_{3}P{O}_4\to C{a}_3(P{O}_4{)}_2+3H_{2}O+3C{O}_2$
The molar masses of $CaC{O}_3,H_{3}P{O}_4$ and $C{a}_3(P{O}_4{)}_2$ are $100$ g/mol, $98$ g/mol and $310$ g/mol respectively.
$50$ g of $CaC{O}_3$ corresponds to $\frac{50}{100}=0.5$ mol.

$70$ g of $H_{3}P{O}_4$ corresponds to $\frac{70}{98}=0.71$ mol.

$3$ moles of $CaC{O}_3$ will react with $2$ moles of $H_{3}P{O}_4$.

$0.5$ moles of $CaC{O}_3$ will react with $\frac{2}{3}\times 0.5=0.33$ moles of $H_{3}P{O}_4$.

However, $0.71$ moles of $H_{3}P{O}_4$ are present.
Hence, $H_{3}P{O}_4$ is the excess reagent and $CaC{O}_3$ is the limiting reagent.
$3$ moles of $CaC{O}_3$ will form $\frac{1}{3}\times 0.5=0.17$ moles of $C{a}_3(P{O}_4{)}_2$.
This corresponds to $0.17\times 310=51.7g$ of $C{a}_3(P{O}_4{)}_2$.
Hence, $0.5$ moles of $CaC{O}_3$ will form $1$ mole of $C{a}_3(P{O}_4{)}_2$.
The amount of excess reagent $H_{3}P{O}_4$ remaining unreacted is $0.71-0.33=0.38$ mol.
It corresponds to $0.38\times 98=37.3$ g.