Question

If 4.50 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(s)+3H2O(g)⟶Fe2O3(s)+3H2(g)

mass:

Answer 1

Given:

P = 0.121 atm

V = 4.5 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 * 4.5 L = n * 0.08206 atm.L/mol.K * 323.2 K

n = 2.053*10^-2 mol

From given chemical equation,

Mol of Fe2O3 formed = (1/3)*mol of H2O reacted

= (1/3)*2.053*10^-2 mol

= 6.84*10^-3 mol

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

use:

mass of Fe2O3,

m = number of mol * molar mass

= 6.84*10^-3 mol * 1.597*10^2 g/mol

= 1.092 g

Answer: 1.09 g