Question

calculations involving a limiting reactant
now consider a situation in which 27.0 g of \\( \ce{p4} \\) is added to 59.0 g of \\( \ce{cl2} \\), and a chemical reaction occurs. to identify the limiting reactant, you will need to perform two separate calculations:

1. calculate the number of moles of \\( \ce{pcl5} \\) that can be produced from 27.0 g of \\( \ce{p4} \\) (and excess \\( \ce{cl2} \\)).
2. calculate the number of moles of \\( \ce{pcl5} \\) that can be produced from 59.0 g of \\( \ce{cl2} \\) (and excess \\( \ce{p4} \\)).

then, compare the two values. the reactant that produces the smaller amount of product is the limiting reactant.

part b

how many moles of \\( \ce{pcl5} \\) can be produced from 27.0 g of \\( \ce{p4} \\) (and excess \\( \ce{cl2} \\))?
express your answer to three significant figures and include the appropriate units.
view available hint(s)

Explanation:

Step1: Write balanced reaction

The reaction is $\text{P}_4 + 10\text{Cl}_2
ightarrow 4\text{PCl}_5$.

Step2: Calculate moles of $\text{P}_4$

Molar mass of $\text{P}_4 = 4 \times 30.97\ \text{g/mol} = 123.88\ \text{g/mol}$.
Moles of $\text{P}_4 = \frac{27.0\ \text{g}}{123.88\ \text{g/mol}} \approx 0.218\ \text{mol}$.

Step3: Relate to moles of $\text{PCl}_5$

From reaction: 1 mol $\text{P}_4$ → 4 mol $\text{PCl}_5$.
Moles of $\text{PCl}_5 = 0.218\ \text{mol} \times 4 = 0.872\ \text{mol}$.

Answer:

0.872 mol