Question

If 8.00 L of water vapor at 50.2 °C and 0.121 atm reacts with excess iron, how many grams of iron(III) oxide will be produced?

2Fe + 3H2O ----- Fe2O3 +3H2

Answer 1

1st find the mol of Fe reacted

we have:

P = 0.121 atm

V = 8.0 L

T = 50.2 oC

= (50.2+273) K

= 323.2 K

find number of moles using:

P * V = n*R*T

0.121 atm * 8 L = n * 0.08206 atm.L/mol.K * 323.2 K

n = 3.65*10^-2 mol

from reaction,

mol of Fe2O3 formed = (1/2)*mol of Fe reacted

= (1/2)*3.65*10^-2 mol

= 1.825*10^-2 mol

Molar mass of Fe2O3 = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

we have below equation to be used:

mass of Fe2O3,

m = number of mol * molar mass

= 1.825*10^-2 mol * 159.7 g/mol

= 2.915 g

Answer: 2.92 g