TOPIC 9. CHEMICAL CALCULATIONS III - stoichiometry.

Ratio: 3 mole 2 mole Moles of Fe available = mass / molar mass = 5.59 / 55.9 = 0.100 mole Moles of O 2 available = mass / molar mass = 1.60 / 32.0 = 0.0500 mole From the ratios above, 1 mole of O 2 requires 1.5 mole of Fe for complete reaction. Moles of O 2 available = 0.0500 which would react with 1.50 × 0.0500 mole of Fe = 0.0750 mole Fe ................
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