Question

13.How many grams of phosphorus (P4) are needed to completely consume 79.2 L of chlorine gas according to the following reaction at 25 °C and 1 atm?

phosphorus (P4) ( s ) + chlorine ( g ) -----> phosphorus trichloride ( l )

Answer 1

73.2g

Step-by-step explanation:

The reaction expression is given as:

P₄ + 6Cl₂ → 4PCl₃

Given parameters:

Volume of chlorine gas = 79.2L

Unknown:

Mass of Phosphorus needed = ?

Solution:

To solve this problem, let us find the number of moles of the chlorine gas.

Since the condition of the reaction is at STP;

22.4L of gas is contained in 1 mole

79.2L of chlorine gas will contain
`(79.2)/(22.4)` = 3.54mole

From the reaction expression;

6 moles of chlorine gas will react with 1 mole of P₄

3.54 mole of chlorine gas will completely react with
`(3.54)/(6)` = 0.59mole of P₄

Mass of P₄ = number of moles x molar mass

Molar mass of P₄ = 4 x 31 = 124g/mol

Mass of P₄ = 0.59 x 124 = 73.2g