Question

phosphorus trifluoride is formed from its elements P4 (s) F2 (g) ---> PF3 (g) how many grams of fluorine are needed to react with 6.20 g

Answer 1

This is an incomplete question, here is a complete question.

Phosphorus trifluoride is formed from its elements.

$P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

How many grams of fluorine are needed to react with 6.20 g of phosphorus?

Answer : The mass of $F_2$ needed are, 11.4 grams.

Explanation : Given,

Mass of $P_4$ = 6.20 g

Molar mass of $P_4$ = 124 g/mol

Molar mass of $F_2$ = 38 g/mol

First we have to calculate the moles of $P_4$

$\text{Moles of }P_4=\frac{\text{Given mass }P_4}{\text{Molar mass }P_4}$

$\text{Moles of }P_4=\frac{6.20g}{124g/mol}=0.05mol$

Now we have to calculate the moles of $F_2$

The balanced chemical equation is:

$P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

From the balanced reaction we conclude that

As, 1 mole of $P_4$ react with 6 moles of $F_2$

So, 0.05 moles of $P_4$ react with $0.05\times 6=0.30$ moles of $F_2$

Now we have to calculate the mass of $F_2$

$\text{ Mass of }F_2=\text{ Moles of }F_2\times \text{ Molar mass of }F_2$

$\text{ Mass of }F_2=(0.30moles)\times (38g/mole)=11.4g$

Therefore, the mass of $F_2$ needed are, 11.4 grams.