Consider the following reaction equation
C₂H_4(g) + 3O_2(g) $\to$ 2CO_2(g) + 2H₂O_(l).

The volume of oxygen at s.t.p that will be required to burn 14g of ethene is
[C₂H₄ = 28; Molar volume of gas at s.t.p = 22.4dm³

64.2dm³

33.6dm³

11.2dm³

3.73dm³

Correct answer is B

C₂H_4(g) + 3O_2(g) $\to$ 2CO_2(g) + 2H₂O_(l)

1 mole of ethene required 3 moles of O₂
28g of ethene required 3 x 22.4dm³ of O₂

14g will require $\frac{14 \times 3 \times 22.4dm^3}{28}$

= 33.6dm³

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Consider the following reaction equation
C₂H_4(g) + 3O_2(g) $\to$ 2CO_2(g) + 2H₂O_(l).

The volume of oxygen at s.t.p that will be required to burn 14g of ethene is
[C₂H₄ = 28; Molar volume of gas at s.t.p = 22.4dm³

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Consider the following reaction equation C2H4(g) + 3O2(g) →\to 2CO2(g) + 2H2O(l). The volume of oxygen at s.t.p that will be required to burn 14g of ethene is [C2H4 = 28; Molar volume of gas at s.t.p = 22.4dm3

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Consider the following reaction equation
C₂H_4(g) + 3O_2(g) $\to$ 2CO_2(g) + 2H₂O_(l).

The volume of oxygen at s.t.p that will be required to burn 14g of ethene is
[C₂H₄ = 28; Molar volume of gas at s.t.p = 22.4dm³