Question

$\text{'}50\text{'}\mathrm{g}$ of ${\mathrm{CaCO}}_3$ is allowed to react with$\text{'}70\text{'}\mathrm{g}$ of ${\mathrm{H}}_3{\mathrm{PO}}_4$_.

Calculate:

(i) Amount of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$formed

(ii) Amount of unreacted reagent

Answer 1

When ${\mathrm{CaCO}}_3$ is allowed to react with ${\mathrm{H}}_3{\mathrm{PO}}_4$ the balanced chemical equation is

$\underset{\mathrm{Phosphoric}\mathrm{acid}}{2{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)}+\underset{\mathrm{Calcium}\mathrm{Carbonate}}{3{\mathrm{CaCO}}_3\left(\mathrm{aq}\right)}\to \underset{\mathrm{Tricalcium}\mathrm{phosphate}}{{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{3{\mathrm{CO}}_2\left(\mathrm{g}\right)}$

Molecular mass of ${\mathrm{CaCO}}_3=100\mathrm{gm}$

Weight of ${\mathrm{CaCO}}_3$(taken) $=50\mathrm{gm}$

Number of moles of ${\mathrm{CaCO}}_3=\frac{\mathrm{Mass}\mathrm{of}\mathrm{substance}\mathrm{given}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{substance}}=\frac{50}{100}=0.5\mathrm{mole}$

The molecular mass of ${\mathrm{H}}_3{\mathrm{PO}}_4=98\mathrm{gm}$

Weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$ (taken)$=70\mathrm{gm}$

Number of moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$= 70/98 = 0.71 mole

The molecular weight of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ (produced) = 310 gm

(i) 3x100 gm of CaCO₃ produce 310 gm of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

1 gm of CaCO₃ will produce $\frac{310}{3\times 100}gm$ ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

50 gm of CaCO₃ will produce $\frac{310\times 50}{3\times 100}gm$ ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

50 gm of CaCO₃ will produce $\mathit{51}\mathit{.}\mathit{66}\mathit{}gm$ ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

(ii) 2x 98 gm ${\mathrm{H}}_3{\mathrm{PO}}_4$ required for 3x100gm of ${\mathrm{CaCO}}_3$

1 gm H₃PO₄ required for $\frac{3\times 100}{2\times 98}gm$${\mathrm{CaCO}}_3$

70 gm H₃PO₄ required for $\frac{3\times 100\times 70}{2\times 98}gm$ ${\mathrm{CaCO}}_3$

70 gm H₃PO₄ required for $\mathit{107}\mathit{.}\mathit{14}\mathit{}gm$ CaCO₃

But given the amount of ${\mathrm{CaCO}}_3$ is less than 107.14 So ${\mathrm{CaCO}}_3$is a limiting reagent.

3x100gm of ${\mathrm{CaCO}}_3$ is required for 2x98 gm ${\mathrm{H}}_3{\mathrm{PO}}_4$

1gm of ${\mathrm{CaCO}}_3$is required for $\frac{2\times 98}{3\times 100}gm$ ${\mathrm{H}}_3{\mathrm{PO}}_4$

50gm of ${\mathrm{CaCO}}_3$ is required for $\frac{2\times 98\times 50}{3\times 100}gm$${\mathrm{H}}_3{\mathrm{PO}}_4$

50gm of ${\mathrm{CaCO}}_3$ is required for $\mathit{32}\mathit{.}\mathit{66}\mathit{}gm$ ${\mathrm{H}}_3{\mathrm{PO}}_4$

Amount of unreacted ${\mathrm{H}}_3{\mathrm{PO}}_4$$\mathit{(}\mathit{70}\mathit{-}\mathit{32}\mathit{.}\mathit{66}\mathit{)}\mathit{}\mathit{=}\mathit{}\mathit{37}\mathit{.}\mathit{33}gm$

Therefore, (i) 51.66 gm of Ca₃(PO₄)₂ formed

(ii) Amount of unreacted reagent(H₃PO₄ )=37.33 gm.