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

Answer:

73.2g

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 $\frac{79.2}{22.4}$   = 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 $\frac{3.54}{6}$   = 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