Question

Consider the following reaction:

P4(s) + 6 H2 (g) ––> 4 PH3 (g)

What mass of P4 will completely react with 2.50 L of hydrogen gas, at 0˚C and 1.50 atm pressure?

Answer 1

To find the mass of P₄ required, we first need to find moles of H₂ that are present. we can use the ideal gas law, PV = nRT

P = pressure = 1.50 atm

V = volume = 2.50 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in$K=0°c+273=273K$

Solving for n (moles) we have

$n=\frac{PV}{RT}=\frac{(1.50)(2.50)}{(0.0821)(273)}=0.167moles H_2$ mass of P4 that will react completely :

$0.167molesH_2×\frac{1molP_4}{6molesH_2}×\frac{123.9gP_4}{molP_4}=3.45gP_4$