Question

If 6.00 L of water vapor at 50.2 °C and 12.26 kPa reacts with excess iron, how many grams of iron(III) oxide are produced?

2Fe(s)+3H2O(g)⟶Fe2O3(s)+3H2(g)

Answer 1

PV = nRT

where, P = pressure = 12.26 kPa = 0.121 atm

V = volume = 6.00 L

n = number of moles

R = Gas constant

T = temperature = 50.2 + 273 = 323.2 K

0.121 * 6.00 = n * 0.0821 * 323.2

0.726 = n * 26.5

n = 0.726 / 26.5 = 0.0274 mole

The number of moles of water vapor = 0.0274

From the balanced equation we can say that

3 mole of H2O produces 1 mole of Fe2O3 so

0.0274 mole of H2O will produce

= 0.0274 mole of H2O *( 1 mole of Fe2O3 / 3 mole of H2O)

= 0.00913 mole of Fe2O3

mass of 1 mole of Fe2O3 = 159.69 g

so the mass of 0.00913 mole of Fe2O3 = 1.46 g

Therefore, the mass of Fe2O3 produced would be 1.46 g