Question

one of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. if 29.8 g of ca3(po4)2 reacts with 54.3 g of h2so4, what is the percent yield if 10.9 g of h3po4 is formed via the unbalanced equation below? ca3(po4)2(s) + h2so4(aq) → h3po4(aq) + caso4(aq)

Explanation:

Step1: Balance the chemical equation

$$\mathrm{Ca_3(PO_4)_2(s)+3H_2SO_4(aq)\to 2H_3PO_4(aq)+3CaSO_4(aq)}$$

Step2: Calculate the molar mass

• Molar mass of $\mathrm{Ca_3(PO_4)_2}$: $3\times40+(31 + 4\times16)\times2=310\ \mathrm{g/mol}$
• Molar mass of $\mathrm{H_2SO_4}$: $2\times1+32 + 4\times16=98\ \mathrm{g/mol}$
• Molar mass of $\mathrm{H_3PO_4}$: $3\times1+31 + 4\times16=98\ \mathrm{g/mol}$

Step3: Calculate the moles of reactants

• Moles of $\mathrm{Ca_3(PO_4)_2}$: $n=\frac{m}{M}=\frac{29.8\ \mathrm{g}}{310\ \mathrm{g/mol}}\approx0.0961\ \mathrm{mol}$
• Moles of $\mathrm{H_2SO_4}$: $n=\frac{m}{M}=\frac{54.3\ \mathrm{g}}{98\ \mathrm{g/mol}}\approx0.554\ \mathrm{mol}$

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of $\mathrm{Ca_3(PO_4)_2}$ to $\mathrm{H_2SO_4}$ is $1:3$.
For $0.0961\ \mathrm{mol}$ of $\mathrm{Ca_3(PO_4)_2}$, the required moles of $\mathrm{H_2SO_4}$ is $0.0961\times3 = 0.2883\ \mathrm{mol}$. Since $0.554\ \mathrm{mol}$ of $\mathrm{H_2SO_4}$ is available (more than required), $\mathrm{Ca_3(PO_4)_2}$ is the limiting reactant.

Step5: Calculate the theoretical yield of $\mathrm{H_3PO_4}$

From the balanced equation, mole ratio of $\mathrm{Ca_3(PO_4)_2}$ to $\mathrm{H_3PO_4}$ is $1:2$.
Moles of $\mathrm{H_3PO_4}$ produced (theoretical) $= 2\times0.0961\ \mathrm{mol}=0.1922\ \mathrm{mol}$
Theoretical mass of $\mathrm{H_3PO_4}$: $m = nM=0.1922\ \mathrm{mol}\times98\ \mathrm{g/mol}\approx18.84\ \mathrm{g}$

Step6: Calculate the percent yield

Percent yield $=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%=\frac{10.9\ \mathrm{g}}{18.84\ \mathrm{g}}\times100\%\approx57.9\%$

Answer:

The percent yield is approximately $57.9\%$