Question

What is the theoretical yield (in grams) of P4O10 if 35.8 g of P4 are reacted with 40.0 g of O2?

Answer 1

Answer on Question #43920 - Chemistry - Inorganic Chemistry

Task:

What is the theoretical yield (in grams) of P4O10 if 35.8 g of P4 are reacted with 40.0 g of O2?

Solution:

1. The chemical equation for the reaction is

$$P_4 + 5O_2 = P_4O_{10}$$

2. The number of moles of $P_4$ is

$$n(P_4) = \frac{m(P_4)}{M(P_4)} = \frac{35.8\ \text{g}}{31\ \frac{\text{g}}{\text{mol}} \cdot 4} = 0.289\ \text{mol}$$

3. The number of moles of $O_2$ is

$$n(O_2) = \frac{m(O_2)}{M(O_2)} = \frac{40.0\ \text{g}}{16\ \frac{\text{g}}{\text{mol}} \cdot 2} = 1.25\ \text{mol}$$

4. According to the equation the ratio of number of moles is $n(P_4) : n(O_2) = 1 : 5$

$n(P_4) = n(O_2)/5 = 1.25/5 = 0.25\ \text{mol}$, but we have more P4 than we need according to the equation. That means that $O_2$ is limiting reagent and we have to use $n(O_2)$ for calculations.

5. The number of moles of $P_4O_{10}$ is

$$n(P_4O_{10}) = \frac{n(O_2)}{5} = \frac{1.25\ \text{mol}}{5} = 0.25\ \text{mol}$$

6. The mass of $P_4O_{10}$ is

$$m(P_4O_{10}) = n(P_4O_{10}) \cdot M(P_4O_{10}) = 0.25 \cdot 284 = 71\ \text{g}$$

Answer:

$$m(P_4O_{10}) = 71\ \text{g}$$

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