Question

Given: P4(s) + 5 O2(g) → P4O10(s), what pressure of O2(g) will be needed at room temperature to completely react with 23.4 g of P4(s) in a 350-mL container

Answer 1

1st find the moles of O2 required

mass of P4 = 23.4 g

molar mass of P4 = 123.88 g/mol

mol of P4 = (mass)/(molar mass)

= 23.4/123.88

= 0.1889 mol

Balanced chemical equation is:

P4(s) + 5 O2(g) → P4O10(s)

According to balanced equation

mol of O2 formed = (5/1)* moles of P4

= (5/1)*0.1889

= 0.9445 mol

Given:

V = 350.0 mL

= (350.0/1000) L

= 0.35 L

n = 0.9445 mol

T = 20.0 oC (room temperature)

= (20.0+273) K

= 293 K

use:

P * V = n*R*T

P * 0.35 L = 0.9445 mol* 0.0821 atm.L/mol.K * 293 K

P = 64.9 atm

Answer: 64.9 atm