Question

How many grams of phosphorus are required to produce 4.21x10^22 molecules of phosphorus trifluoride?

Answer 1

2.1618 grams of P4 is required to produce 4.21x10^22 molecules of phosphorus trifluoride.

Explanation:

The balanced chemical reaction for formation of PF3 from yellow phosphorus is given by:

P4 + 6F2 ⇒ 4PF3

1 mole of P4 reacts to give 4 moles of PF3

It is given that 4.21x10^22 molecules of PF3 are produced

so the number of moles can be calculated by the relation

number of molecules = number of moles × Avagadro number

number of moles = $\frac{number of molecules}{avagadro number}$

n = $\frac{4.21x10^22}{6.023. 10^23}$

n= 0.06989 moles of PF3 is formed.

Applying stoichiometry,

1 mole of P4 gives 4 moles PF3

x mole will produce 0.06989 moles of PF3

$\frac{4}{1}$ = $\frac{0.0698}{x}$

4x= 0.0698 × 1

 x =  $\frac{0.0698}{4}$

x= 0.01745 moles of P4 will be required.

The weight of the phosphorus can be obtained by number of moles × atomic mass of one mole of P4

= 0.01745 × 123.89

= 2.1618 grams of P4.