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In the reaction 2SO3(g) -> 2SO2(g) + O2(g) with 50% yield, what is the minimum mass of SO3(g) required to produce at least 6.4 kg of each product?

1. 64 kg
2. 32 kg
3. 16 kg
4. 8 kg

Correct answer: 64 kg

Solution

From 2 mol SO3, we get 2 mol SO2 and 1 mol O2. For 6.4 kg SO2 (100 mol), we need 100 mol SO3 at 100% yield. For 6.4 kg O2 (200 mol), we need 400 mol SO3 at 100% yield. The limiting requirement is O2 production: 400 mol SO3 = 400*80 g = 32,000 g = 32 kg at 100% yield. At 50% yield, we need 64 kg of SO3.

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