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How many liters of oxygen gas at STP are needed to react completely with 8.0 of magnesium?
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1. Write the balanced chemical equation. 2Mg + O2 → 2MgO 2. Convert grams to moles of magnesium. \[8/24.3 = .33 mol\] 3. Use moles of magnesium to find moles of oxygen. \[\frac{ 1 mol O2 }{ 2 mol Mg } = \frac{ x }{ .33 mol Mg}\] \[x = \frac{ .33 }{ 2 } = .166 mol O2\] 5. One mole of an ideal gas will occupy a volume of 22.4 liters at STP. If you know one mole of O2 takes up 22.4 liters, you can find how much 4 moles will take up. \[\frac{ 1 mol }{ 22.4 L } = \frac{ .166 mol }{ x }\] 6. Solve for x (liters) \[x = (.166)(22.4 L) = 3.7 L\] Answer: 3.7 liters O2
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