Question

one of the commercial uses of sulfuric acid is in the production of calcium sulfate and phosphoric acid. if 23.3 g of ca3(po4)2 reacts with an excess of h2so4, what is the percent yield if 13.4 g of h3po4 are formed in the following unbalanced chemical equation?
ca3(po4)2 (s) + h2so4 (aq) → h3po4 (aq) + caso4 (aq)

Explanation:

Step1: Balance the chemical equation

$$\ce{Ca3(PO4)2(s) + 3H2SO4(aq) \to 2H3PO4(aq) + 3CaSO4(aq)}$$

Step2: Calculate the molar mass of $\ce{Ca3(PO4)2}$

The molar mass of $\ce{Ca}$ is $40.08\ g/mol$, of $\ce{P}$ is $30.97\ g/mol$, and of $\ce{O}$ is $16.00\ g/mol$.
For $\ce{Ca3(PO4)2}$:

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Step3: Calculate the moles of $\ce{Ca3(PO4)2}$

Using the formula $n=\frac{m}{M}$, where $m = 23.3\ g$ and $M = 310.18\ g/mol$

$$n(\ce{Ca3(PO4)2})=\frac{23.3\ g}{310.18\ g/mol}\approx0.0751\ mol$$

Step4: Determine the moles of $\ce{H3PO4}$ produced based on stoichiometry

From the balanced equation, the mole ratio of $\ce{Ca3(PO4)2}: \ce{H3PO4}=1:2$
So, $n(\ce{H3PO4}) = 2\times n(\ce{Ca3(PO4)2})=2\times0.0751\ mol = 0.1502\ mol$

Step5: Calculate the molar mass of $\ce{H3PO4}$

The molar mass of $\ce{H}$ is $1.01\ g/mol$, of $\ce{P}$ is $30.97\ g/mol$, and of $\ce{O}$ is $16.00\ g/mol$

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Step6: Calculate the theoretical yield of $\ce{H3PO4}$

Using $m = n\times M$, with $n = 0.1502\ mol$ and $M = 98.00\ g/mol$

$$m_{theoretical}=0.1502\ mol\times98.00\ g/mol = 14.72\ g$$

Step7: Calculate the percent yield

Using the formula $\text{Percent Yield}=\frac{m_{actual}}{m_{theoretical}}\times100\%$, where $m_{actual}=13.4\ g$ and $m_{theoretical}=14.72\ g$

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Answer:

The percent yield is approximately $91.0\%$