If 66.2 Liters of Carbon Monoxide gas react with excess Hydrogen, how many grams of methanol (CH3OH) can be made? CO (g) + 2 H2 (g) ⟶ CH3OH (g)

Question

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Answer 1

CO (g) + 2 H2 (g) ⟶ CH3OH (g)

66.2 L CO x 1 L CH3OH/1 L CO = 66.2 L CH3OH formed (based on 1:1 ratio of CO : CH3OH in balanced equation)

Assuming STP, 66.2 L CH3OH x 1 mole/22.4 L = 2.955 moles CH3OH

Mass CH3OH = 2.955 moles x 32g/mole = 94.6 g (3 sig. figs.)