Question

50 g of CaCO3 is allowed to react with 73.5 g of H3PO4. $CaCO_3 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O + CO_2$ Calculate : (i) Amount of $Ca_3(PO_4)_2$ formed (in moles) (ii) Amount of unreacted reagent (in moles)

Answer 1

(i) 1/6 mol, (ii) 5/12 mol

Answer 2

Solution:

The reaction between $CaCO_3$ and $H_3PO_4$ is given by the unbalanced equation: $CaCO_3 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O + CO_2$

Step 1: Balance the chemical equation.

The balanced equation is: $3CaCO_3 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 3H_2O + 3CO_2$

Step 2: Calculate the number of moles of each reactant.

Molar mass of $CaCO_3 = 40.08 + 12.01 + 3 \times 16.00 = 100.09$ g/mol. Using approximate values (Ca=40, C=12, O=16), Molar mass = $40 + 12 + 3 \times 16 = 100$ g/mol. Molar mass of $H_3PO_4 = 3 \times 1.01 + 30.97 + 4 \times 16.00 = 98.00$ g/mol. Using approximate values (H=1, P=31, O=16), Molar mass = $3 \times 1 + 31 + 4 \times 16 = 98$ g/mol.

Moles of $CaCO_3 = \frac{\text{Mass of } CaCO_3}{\text{Molar mass of } CaCO_3} = \frac{50 \text{ g}}{100 \text{ g/mol}} = 0.5 \text{ mol}$. Moles of $H_3PO_4 = \frac{\text{Mass of } H_3PO_4}{\text{Molar mass of } H_3PO_4} = \frac{73.5 \text{ g}}{98 \text{ g/mol}} = 0.75 \text{ mol}$.

Step 3: Identify the limiting reagent.

According to the balanced equation, 3 moles of $CaCO_3$ react with 2 moles of $H_3PO_4$. The stoichiometric ratio is $n(CaCO_3) : n(H_3PO_4) = 3 : 2 = 1.5$. The available mole ratio is $n(CaCO_3) : n(H_3PO_4) = 0.5 : 0.75 = \frac{0.5}{0.75} = \frac{50}{75} = \frac{2}{3} \approx 0.667$. Since the available ratio (0.667) is less than the required stoichiometric ratio (1.5), $CaCO_3$ is the limiting reagent.

Alternatively, we can check how much of one reactant is required to react with the given amount of the other. Moles of $H_3PO_4$ required to react with 0.5 mol of $CaCO_3 = 0.5 \text{ mol } CaCO_3 \times \frac{2 \text{ mol } H_3PO_4}{3 \text{ mol } CaCO_3} = \frac{1}{3} \text{ mol } H_3PO_4$. Since the available amount of $H_3PO_4$ (0.75 mol) is greater than the required amount (1/3 mol), $H_3PO_4$ is in excess, and $CaCO_3$ is the limiting reagent.

(i) Calculate the amount of $Ca_3(PO_4)_2$ formed (in moles).

The amount of product formed is determined by the limiting reagent ($CaCO_3$). 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 = Moles of $CaCO_3$ used $\times \frac{1 \text{ mol } Ca_3(PO_4)_2}{3 \text{ mol } CaCO_3}$ Moles of $Ca_3(PO_4)_2$ formed = $0.5 \text{ mol} \times \frac{1}{3} = \frac{0.5}{3} = \frac{1}{6} \text{ mol}$.

(ii) Calculate the amount of unreacted reagent (in moles).

The unreacted reagent is the excess reagent, which is $H_3PO_4$. Amount of $H_3PO_4$ reacted with 0.5 mol of $CaCO_3 = \frac{1}{3}$ mol (calculated in Step 3). Initial amount of $H_3PO_4 = 0.75 \text{ mol} = \frac{3}{4} \text{ mol}$. Amount of $H_3PO_4$ unreacted = Initial amount of $H_3PO_4$ - Amount of $H_3PO_4$ reacted Amount of $H_3PO_4$ unreacted = $\frac{3}{4} \text{ mol} - \frac{1}{3} \text{ mol} = \frac{9}{12} \text{ mol} - \frac{4}{12} \text{ mol} = \frac{5}{12} \text{ mol}$.