# What volume of hydrogen is required to interact with iron (III) oxide weighing 640g,

What volume of hydrogen is required to interact with iron (III) oxide weighing 640g, containing 25% impurities? What mass of water is formed in this case?

1. Let’s compose the reaction of chemical interaction.
Fe2O3 + 3H2 = 2Fe + 3H2O.
2. Find the amount of reacted iron oxide (III).
ω (Fe2O3) = 100% – ω (impurities) = 100% – 25% = 75%.
m (Fe2O3) = m (mixture) * ω (Fe2O3) / 100% = 640 g * 75% / 100% = 480 g.
n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 480 g / 160 g / mol = 3 mol.
3. Using the reaction equation, we find the amount and then the required volume of hydrogen.
n (H2) = n (Fe2O3) * 3 = 3 mol * 3 = 9 mol.
V (H2) = n (H2) * Vm = 9 mol * 22.4 l / mol = 201.6 l.
Answer: V (H2) = 201.6 liters.

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