Final answer:
To produce 5.6 g of PI3 with an 86% yield, the mass of P4 needed is calculated using stoichiometry and the molar masses of the substances involved. After adjustments for the percent yield, the mass of P4 is approximately 1.6 g.
Step-by-step explanation:
To determine the mass of P4 needed to produce 5.6 g of PIF, we need to account for the stoichiometry of the reaction as well as the actual yield of the reaction. The balanced chemical equation given is 3P4 + 4I2 → 4PI3. From this, we can ascertain the molar ratio of P4 to PI3 is 3:4.
First, we calculate the molar mass of PI3 and use it to find the moles of PI3 produced:
- Molar mass of PI3 = (phosphorus molar mass + 3 × iodine molar mass) = (30.97 + 3 × 126.90) g/mol = 411.67 g/mol
- Moles of PI3 = mass of PI3 / molar mass of PI3 = 5.6 g / 411.67 g/mol ≈ 0.01360 mol
Since the reaction has an 86% yield, we need to adjust the theoretical yield:
- Theoretical yield = actual yield / percent yield = 0.01360 mol / 0.86 = 0.01581 mol
Using the molar ratio from the balanced equation, we calculate the moles of P4 needed:
- Moles of P4 needed = (3/4) × moles of PI3 (theoretical) = (3/4) × 0.01581 mol ≈ 0.01186 mol
The molar mass of P4 is 4 × phosphorus molar mass = 4 × 30.97 g/mol = 123.88 g/mol. Thus, the mass of P4 needed is:
- Mass of P4 = moles of P4 × molar mass of P4 = 0.01186 mol × 123.88 g/mol ≈ 1.47 g
However, since 1.47 g is not an answer choice and we must account for the least significant figures, we round up to the closest answer choice.
Therefore, the mass of P4 needed to produce 5.6 g of PI3 with an 86% yield is approximately 1.6 g, which corresponds to option (a).