Question

Phosphorus \(\left(P_{4}\right)\) is commercially prepared by heating a mixture of calcium phosphate (CaSiO \(_{3} ),\) sand \(\left(\mathrm{SiO}_{2}\right)\) and coke (C) in an electric furnace. The process involves two reactions. $$ \begin{array}{c}{2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+6 \mathrm{SiO}_{2}(\mathrm{s}) \rightarrow 6 \mathrm{CaSiO}_{3}(1)+\mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{g})} \\\ {\mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{g})+10 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{P}_{4}(\mathrm{g})+10 \mathrm{CO}(\mathrm{g})}\end{array} $$ The \(\mathrm{P}_{4} \mathrm{O}_{10}\) produced in the first reaction reacts with an excess of coke (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 \(\mathrm{P}_{4}\) .

Short answer

The theoretical yield of P4 is 49.93 g, and the percent yield of P4 is 90.1 % based on the actual yield of 45.0 g.

Step-by-step solution

Convert Given Masses to Moles

To convert the given masses to moles, we need the molar masses of calcium phosphate (Ca3(PO4)2) and sand (SiO2). Molar mass of Ca3(PO4)2 = 3(40.08) + 2(94.97) = 310.18 g/mol, and molar mass of SiO2 = 60.08 g/mol. Now, we can calculate the moles of Ca3(PO4)2 and SiO2. Moles of Ca3(PO4)2 = (250.0 g) / (310.18 g/mol) = 0.8062 mol Moles of SiO2 = (400.0 g) / (60.08 g/mol) = 6.659 mol

Find Limiting Reactant

Now we need to find the limiting reactant. We can do this by comparing the mole ratios of Ca3(PO4)2 and SiO2 from the balanced reaction. For every 2 moles of Ca3(PO4)2, there are 6 moles of SiO2 consumed. Therefore, the mole ratio of Ca3(PO4)2: SiO2 = 2:6 or 1:3. Dividing the moles of each substance by their stoichiometric coefficient, we have: Moles of Ca3(PO4)2 = 0.8062 mol / 2 = 0.4031 Moles of SiO2 = 6.659 mol / 6 = 1.1098 Since the value for Ca3(PO4)2 (0.4031) is smaller than SiO2 (1.1098), Ca3(PO4)2 is the limiting reactant.

Calculate Moles of P4 Produced

Using the stoichiometry of the reaction, we can calculate the moles of P4 produced by the limiting reactant (Ca3(PO4)2). For every 2 moles of Ca3(PO4)2 reacted, 1 mole of P4O10 is produced, and 1 mole of P4O10 reacts with 10 moles of coke (C) to produce 1 mole of P4. Thus, the moles of P4 produced can be calculated as follows: Moles of P4 = moles of Ca3(PO4)2 * (1 mol P4O10 / 2 mol Ca3(PO4)2) * (1 mol P4 / 1 mol P4O10) = 0.8062 * (1/2) = 0.4031 mol

Convert Moles of P4 to Grams

Now we can convert the moles of P4 produced to grams. The molar mass of P4 is 4(30.97) = 123.88 g/mol. Theoretical Yield of P4 = (0.4031 mol) * (123.88 g/mol) = 49.93 g

Calculate Percent Yield

Now we can calculate the percent yield of P4 using the actual yield (45.0 g P4) and theoretical yield (49.93 g P4). Percent Yield = (actual yield / theoretical yield) * 100 Percent Yield = (45.0 g / 49.93 g) * 100 = 90.1 % Therefore, the percent yield of phosphorus is 90.1%.

Key concepts

Theoretical Yield

Understanding theoretical yield is fundamental in the study of chemical reactions. It refers to the maximum amount of product that could be generated from a given amount of reactants, assuming the reaction goes to completion without any losses. It is a perfect scenario where every molecule reacts exactly as outlined in the stoichiometric equation.

To calculate the theoretical yield, one must first identify the limiting reactant—the substance that will be completely consumed first in the reaction, thus dictating the maximum amount of product formed. Once the limiting reactant is determined, stoichiometry is used to convert the number of moles of the limiting reactant to moles of the desired product, which can then be converted to grams using the molar mass of the product. This result represents the theoretical yield, an ideal quantity that guides our expectations for the reaction in a perfect world.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction and is defined as the ratio of actual yield to theoretical yield, multiplied by 100. The actual yield is the quantity of product that is actually obtained from a reaction, which is often less than the theoretical yield due to factors such as incomplete reactions, side reactions, and loss of product during the collection process.

To find the percent yield, you simply divide the actual yield by the theoretical yield and then multiply that result by 100. For instance, if in a certain reaction the actual yield of product was 45.0 grams but the theoretical yield was calculated to be 49.93 grams, the percent yield would be \( (45.0 \, \mathrm{g} / 49.93 \, \mathrm{g}) \times 100 = 90.1 \% \). This value serves as a practical measure of the reaction's success, allowing chemists to assess the effectiveness of their methodologies and optimize processes for greater yield.

Limiting Reactant

The concept of the limiting reactant is pivotal in stoichiometry, as it determines the amount of product that can be formed in a chemical reaction. The limiting reactant is the reactant that will be exhausted first, thus limiting the extent of the reaction and the amount of product that can be formed. When performing stoichiometric calculations, identifying the limiting reactant is crucial since it directly affects the calculation of the theoretical yield.

In a balanced chemical equation, we can compare the molar ratios of reactants to determine which reactant will limit the reaction. To identify the limiting reactant, we calculate the moles of each reactant present and then divide those values by their respective coefficients in the balanced equation. Whichever reactant has the smallest quotient is the limiting reactant. This is an essential step in problem-solving where multiple reactants are involved, as seen in the textbook exercise where calcium phosphate (Ca₃(PO₄)₂) is the limiting reactant when mixed with sand (SiO₂).