A sample of 6.20 g of white phosphorus (P₄, 124.0 g/mol) burns in 10.65 g of chlorine (71.0 g/mol)…

Step 1. Moles. $\text{P}_{4}$: $\dfrac{6.20}{124.0} = 0.0500\ \text{mol}$. $\text{Cl}_{2}$: $\dfrac{10.65}{71.0} = 0.150\ \text{mol}$.

Step 2. The limiting reagent. The equation is $\text{P}_{4}\ +\ 6\text{Cl}_{2}\ \rightarrow \ 4\text{PCl}_{3}$. For 0.0500 mol of $\text{P}_{4}$ we need $6 \times 0.0500 = 0.300\ \text{mol}$ of chlorine. There is only 0.150 mol, so chlorine limits.

Step 3. Trichloride. $0.150\ \text{mol of Cl}_2 \times \dfrac{4}{6} = 0.100\ \text{mol}$ of $\text{PCl}_{3}$. The mass is $0.100 \times 137.5 = 13.75\ \text{g}$.

Step 4. Phosphorus left. The reaction uses $\dfrac{0.150}{6} = 0.0250\ \text{mol}$ of $\text{P}_{4}$. The rest is $0.0500 - 0.0250 = 0.0250\ \text{mol}$, which is $0.0250 \times 124.0 = 3.10\ \text{g}$.

Step 5. Pentachloride. With an excess of chlorine, $\text{P}_{4}\ +\ 10\text{Cl}_{2}\ \rightarrow \ 4\text{PCl}_{5}$ uses all the phosphorus: $4 \times 0.0500 = 0.200\ \text{mol}$. The mass is $0.200 \times 208.5 = 41.7\ \text{g}$.