Question

50 gm of $CaCO_3$ is allowed to react with 68.6 gm of $H_3PO_4$ then select the correct option(s)- $3CaCO_3 + 2H_3PO_4 \longrightarrow Ca_3(PO_4)_2 + 3H_2O + 3CO_2$

• A. 51.67 gm salt is formed
• B. Amount of unreacted reagent = 35.93 gm
• C. $n_{CO2}$ = 0.5 moles
• D. 0.7 mole $CO_2$ is evolved

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The problem involves a stoichiometry calculation for the reaction between calcium carbonate ($CaCO_3$) and phosphoric acid ($H_3PO_4$).

1. Balanced Chemical Equation: The given balanced chemical equation is: $3CaCO_3 + 2H_3PO_4 \longrightarrow Ca_3(PO_4)_2 + 3H_2O + 3CO_2$

2. Molar Masses of Reactants and Products:

• Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100$ g/mol
• Molar mass of $H_3PO_4 = (3 \times 1) + 31 + (4 \times 16) = 3 + 31 + 64 = 98$ g/mol
• Molar mass of $Ca_3(PO_4)_2 = (3 \times 40) + 2 \times (31 + (4 \times 16)) = 120 + 2 \times (31 + 64) = 120 + 2 \times 95 = 120 + 190 = 310$ g/mol
• Molar mass of $CO_2 = 12 + (2 \times 16) = 12 + 32 = 44$ g/mol

3. Calculate Initial Moles of Reactants:

• Moles of $CaCO_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{50 \text{ gm}}{100 \text{ gm/mol}} = 0.5 \text{ mol}$
• Moles of $H_3PO_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{68.6 \text{ gm}}{98 \text{ gm/mol}} = 0.7 \text{ mol}$

4. Identify the Limiting Reagent: To find the limiting reagent, divide the moles of each reactant by its stoichiometric coefficient in the balanced equation:

• For $CaCO_3$: $\frac{0.5 \text{ mol}}{3} = 0.1667$
• For $H_3PO_4$: $\frac{0.7 \text{ mol}}{2} = 0.35$

Since $0.1667 < 0.35$, $CaCO_3$ is the limiting reagent. It will be completely consumed, and all calculations for products and unreacted reagent will be based on the amount of $CaCO_3$.

5. Evaluate Each Option:

(A) 51.67 gm salt ($Ca_3(PO_4)_2$) is formed: From the balanced equation, 3 moles of $CaCO_3$ produce 1 mole of $Ca_3(PO_4)_2$. Moles of $Ca_3(PO_4)_2$ formed = $0.5 \text{ mol } CaCO_3 \times \frac{1 \text{ mol } Ca_3(PO_4)_2}{3 \text{ mol } CaCO_3} = \frac{0.5}{3} = 0.1666... \text{ mol}$ Mass of $Ca_3(PO_4)_2$ formed = $0.1666... \text{ mol} \times 310 \text{ gm/mol} = 51.666... \text{ gm} \approx 51.67 \text{ gm}$ So, option (A) is correct.

(B) Amount of unreacted reagent = 35.93 gm: The unreacted reagent is $H_3PO_4$. From the balanced equation, 3 moles of $CaCO_3$ react with 2 moles of $H_3PO_4$. Moles of $H_3PO_4$ reacted = $0.5 \text{ mol } CaCO_3 \times \frac{2 \text{ mol } H_3PO_4}{3 \text{ mol } CaCO_3} = \frac{1}{3} = 0.3333... \text{ mol}$ Moles of $H_3PO_4$ unreacted = Initial moles - Moles reacted $= 0.7 \text{ mol} - 0.3333... \text{ mol} = 0.3666... \text{ mol}$ Mass of unreacted $H_3PO_4$ = $0.3666... \text{ mol} \times 98 \text{ gm/mol} = 35.9333... \text{ gm} \approx 35.93 \text{ gm}$ So, option (B) is correct.

(C) $n_{CO2}$ = 0.5 moles: From the balanced equation, 3 moles of $CaCO_3$ produce 3 moles of $CO_2$. Moles of $CO_2$ produced = $0.5 \text{ mol } CaCO_3 \times \frac{3 \text{ mol } CO_2}{3 \text{ mol } CaCO_3} = 0.5 \text{ mol}$ So, option (C) is correct.

(D) 0.7 mole $CO_2$ is evolved: As calculated above, 0.5 moles of $CO_2$ are evolved, not 0.7 moles. So, option (D) is incorrect.

Conclusion: Options (A), (B), and (C) are correct.

Explanation of the solution:

1. Calculate initial moles of reactants using their given masses and molar masses.
2. Identify the limiting reagent by comparing the mole ratios of reactants to their stoichiometric coefficients. $CaCO_3$ is the limiting reagent.
3. Use the moles of the limiting reagent to calculate the moles and mass of the product $Ca_3(PO_4)_2$ formed.
4. Use the moles of the limiting reagent to calculate the moles of $CO_2$ evolved.
5. Use the moles of the limiting reagent to calculate the moles of $H_3PO_4$ reacted, then subtract from initial moles to find unreacted amount.