50 g of ${C a C O}_{3}$ is allowed to react with 70 g of $H_{3} P O_{4}$ . Calculate :
(i) amount of ${C a}_{3} ({P O}_{4})_{2}$ formed
(ii) amount of unreacted reagent

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Solution

$2 : 3 : 1 {2 H}_{3} {P O}_{4} + 3 C a C O_{3} ⟶ {C a}_{3} (P O_{4})_{2} + 3 H_{2} 0 + 3 C O_{2} 1 m o l = 98 g 1 m o l = 100 g 70 g 50 g = \frac{5}{7} m o l = \frac{1}{2} m o l F o r \frac{1}{2} m o l C a {C o}_{3}, w e n e e d \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} m o l H_{3} {P O}_{4} ∴ {C a}_{3} ({P O}_{4})_{2} f o r m e d = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} m o l e s = \frac{1}{6} \times 310 g = 51.66 g u n r e a c t e d H_{3} {P O}_{4} = \frac{5}{7} - \frac{1}{3} = \frac{8}{21} m o l e s = \frac{8}{21} \times 98 = 37.33 g$

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