Problem 96 - Chemistry Matter and Change

Question

Phosphorus (P. ) is commercially prepared by heating a mixture of calcium phosphate \(\left(\mathrm{Ca} \mathrm{SiO}_{3}\right),\) sand \(\left(\mathrm{SiO}_{2}\right)\) and coke \((\mathrm{C})\) in an electric furnace. The process involves two reactions. \begin{equation} 2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+6 \mathrm{SiO}_{2}(\mathrm{s}) \rightarrow 6 \mathrm{CaSiO}_{3}(\mathrm{l})+\mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{g}) \end{equation} \begin{equation} \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{g})+10 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{P}_{4}(\mathrm{g})+10 \mathrm{CO}(\mathrm{g}) \end{equation} The P \(_{4} \mathrm{O}_{10}\) produced in the first reaction reacts with an excess of coke \((\mathrm{C})\) in the second reaction. Determine the theoretical yield of \(\mathrm{P}_{4}\) if 250.0 \(\mathrm{g}\) of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) and 400.0 \(\mathrm{g}\) of \(\mathrm{SiO}_{2}\) are heated. If the actual yield of \(\mathrm{P}_{4}\) is \(45.0 \mathrm{g},\) determine the percent yield of \(\mathrm{P}_{4} .\)

Step-by-Step Solution

Answer

Theoretical yield of \( P_4 \) is 49.94 g, and the percent yield is 90.1%.

Step 1: Molar Mass Calculation

First, calculate the molar mass of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) and \( \mathrm{SiO}_{2} \). The molar mass of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \): \( 3(40.08) + 2(30.97 + 16\times4) = 310.18 \, \mathrm{g/mol} \). The molar mass of \( \mathrm{SiO}_{2} \): \( 30.97 + 2(16) = 60.08 \, \mathrm{g/mol} \).

Step 2: Convert Mass to Moles

Convert the given masses to moles. For \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \): \( \frac{250.0 \, \mathrm{g}}{310.18 \, \mathrm{g/mol}} \approx 0.806 \, \mathrm{mol} \). For \( \mathrm{SiO}_{2} \): \( \frac{400.0 \, \mathrm{g}}{60.08 \, \mathrm{g/mol}} \approx 6.657 \, \mathrm{mol} \).

Step 3: Determine the Limiting Reagent

According to the reaction, 2 moles of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) react with 6 moles of \( \mathrm{SiO}_{2} \). Therefore, the ratio required is 1:3. For \(0.806 \, \mathrm{mol} \) of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \), \( 0.806 \times 3 = 2.418 \, \mathrm{mol} \) of \( \mathrm{SiO}_{2} \) is needed. Since \( 6.657 \, \mathrm{mol} \) of \( \mathrm{SiO}_{2} \) is available, \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) is the limiting reagent.

Step 4: Calculate Theoretical Yield of \( \mathrm{P}_{4} \)

From the balanced reaction, 2 moles of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) produce 1 mole of \( \mathrm{P}_{4} \). Therefore, \( 0.806 \times \frac{1}{2} = 0.403 \, \mathrm{mol} \) of \( \mathrm{P}_{4} \) can be formed. The molar mass of \( \mathrm{P}_{4} \) is \( 4 \times 30.97 = 123.88 \, \mathrm{g/mol} \). The theoretical yield is \( 0.403 \, \mathrm{mol} \times 123.88 \, \mathrm{g/mol} \approx 49.94 \, \mathrm{g} \).

Step 5: Calculate Percent Yield

Use the formula for percent yield: \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \). The actual yield of \( \mathrm{P}_{4} \) is given as \( 45.0 \, \mathrm{g} \). Thus, percent yield is \( \frac{45.0 \, \mathrm{g}}{49.94 \, \mathrm{g}} \times 100 \approx 90.1\% \).

Key Concepts

Molar Mass Calculation

Understanding molar mass calculation is crucial in stoichiometry. To find the molar mass, you need to add up all the atomic masses of the elements in a compound. For example, the molar mass of calcium phosphate, \( \mathrm{Ca}_3(\mathrm{PO}_4)_2 \), is calculated by summing the masses of its constituent atoms: calcium \((\mathrm{Ca})\), phosphorus \((\mathrm{P})\), and oxygen \((\mathrm{O})\). This compound has three calcium atoms and two phosphate groups, each consisting of one phosphorus atom and four oxygens. So, it is calculated as:

• The mass of three calcium atoms: \(3 \times 40.08\, \mathrm{g/mol} = 120.24\, \mathrm{g/mol}\).
• The mass of two phosphate groups: \(2 \times (30.97 + 4 \times 16\, \mathrm{g/mol}) = 2 \times 94.97\, \mathrm{g/mol} = 189.94\, \mathrm{g/mol}\).
• Add them up: \(310.18\, \mathrm{g/mol}\).

For \(\mathrm{SiO}_2\), you simply add the mass of one silicon atom \(30.97\, \mathrm{g/mol}\) and two oxygen atoms \(2 \times 16\, \mathrm{g/mol}\), giving \(60.08\, \mathrm{g/mol}\). This helps to convert the mass of compounds into moles for reaction calculations.

Theoretical Yield

The theoretical yield is the amount of product that could be formed from a reaction, if everything goes perfectly according to stoichiometry. Once you know the limiting reagent, you can calculate how much product to expect. For the reaction involving \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\) and \(\mathrm{SiO}_2\) to produce \(\mathrm{P}_4\), the balanced chemical equation dictates the mole-to-mole conversion. From the equation, \[2 \mathrm{Ca}_3(\mathrm{PO}_4)_2 + 6 \mathrm{SiO}_2 \rightarrow 6 \mathrm{CaSiO}_3 + \mathrm{P}_4\],\ it can be seen that 2 moles of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\) produce 1 mole of \(\mathrm{P}_4\). This ratio allows us to find that from 0.806 moles of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\), you can theoretically produce:

• \(0.806 \times \frac{1}{2} = 0.403\, \mathrm{mol}\) of \(\mathrm{P}_4\).

By multiplying this by the molar mass of \(\mathrm{P}_4\) (123.88 g/mol), you find the theoretical yield in grams: \(49.94\, \mathrm{g}\). This gives you the amount of \(\mathrm{P}_4\) you could get if everything reacts perfectly.

Limiting Reagent

In every chemical reaction, the limiting reagent is the one that is completely consumed first, thus stopping the reaction from producing more product. To identify the limiting reagent, compare the amount of each reactant to the stoichiometric ratios in the balanced chemical equation. In our example, the equation indicates that \(2\) moles of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\) react with \(6\) moles of \(\mathrm{SiO}_2\), a \(1:3\) ratio. With \(0.806\) moles of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\), you would need \(0.806 \times 3 = 2.418\, \mathrm{mol}\) of \(\mathrm{SiO}_2\). Since you have \(6.657 \mathrm{mol}\) available, which is more than needed, \(\mathrm{SiO}_2\) is in excess. Thus, \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\) is the limiting reagent. Understanding which reagent is limiting allows you to calculate how much product can actually form.

Percent Yield

Percent yield is a measure of the efficiency of a reaction, comparing the actual amount of product obtained to the theoretical amount that could have been produced. It is expressed as a percentage and calculated using the formula:

• Percent Yield = \(\left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\).

In our scenario, the actual yield of \(\mathrm{P}_4\) was \(45.0\, \mathrm{g}\), while the theoretical yield was \(49.94\, \mathrm{g}\). Substituting these values into the formula gives:

• \(\frac{45.0}{49.94} \times 100 = 90.1\%\).

This indicates that the reaction was quite efficient, achieving 90.1% of the possible \(\mathrm{P}_4\) yield. Knowing the percent yield helps in evaluating the performance of chemical reactions and optimizing reaction conditions.