$$P_4 + 6F_2 → 4PF_3$$
From the stoichiometry of the reaction, we can see that 4 moles of phosphorus react with 6 moles of fluorine. The molar mass of phosphorus is 30.97 g/mol, and the molar mass of fluorine is 18.99 g/mol.
Therefore, the number of moles of phosphorus in 6.20 g is:
$$6.20 \text{ g }P \times \frac{1 \text{ mol }P}{30.97 \text{ g }P} = 0.200 \text{ mol }P$$
The number of moles of fluorine needed to react with 0.200 mol of phosphorus is:
$$0.200 \text{ mol }P \times \frac{6 \text{ mol }F_2}{4 \text{ mol }P} = 0.300 \text{ mol }F_2$$
The mass of fluorine needed is:
$$0.300 \text{ mol }F_2 \times 18.99 \text{ g/mol }F_2 = 5.697 \text{ g }F_2$$
Therefore, 5.697 g of fluorine are needed to react with 6.20 g of phosphorus.
