Question

Oxygen gas reacts with powdered iron according to the reaction: 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s). What mass of Fe is required to completely react with 200.0 L of oxygen gas measured at 1055 mmHg and 71.2 °C?

Answer 1

1st find the number of moles of O2

we have:

P = 1055.0 mm Hg

= (1055.0/760) atm

= 1.3882 atm

V = 200.0 L

T = 71.2 oC

= (71.2+273) K

= 344.2 K

find number of moles using:

P * V = n*R*T

1.3882 atm * 200 L = n * 0.08206 atm.L/mol.K * 344.2 K

n = 9.829 mol

This is number of moles of O2

from reaction,

moles of Fe = (4/3)*moles of O2

= (4/3)*9.829 mol

= 13.105 mol

Molar mass of Fe = 55.85 g/mol

we have below equation to be used:

mass of Fe,

m = number of mol * molar mass

= 13.11 mol * 55.85 g/mol

= 732 g

Answer: 732 g