Mole ConceptHard

Question

When 12 g graphite is burnt in sufficient oxygen, CO as well as CO2 is formed. If the product contains 40% CO and 60% CO2 by mass and none of the reactant is left, then what is the mass of oxygen gas used in combustion?

Options

A.24.0 g
B.21.33 g
C.23.8 g
D.15.6 g

Solution

$C + \frac{1}{2}O_{2} \rightarrow CO$

x mole 0.5x mole x × 28 g

C + O2 → CO2

(1 − x) mole (1 − x) mole (1 − x) × 44 g

From question, [28x + 44(1 − x)] × $\frac{40}{100}$= 28x ⇒ x = 0.5116

∴ Mass of O2 used = [0.5x + (1 − x)] × 32 = 23.8 g

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