The amount of carbon dioxide produced when 40 g of methane reacts with 48 g of oxygen is.

44 g

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33 g

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57 g

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Solution

The correct option is B 33 g
Gram molecular mass of methane = 16 g
Gram molecular mass of oxygen = 32 g

The balanced chemical reaction between methane and oxygen is as follows:
${CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (g) 16 g 64 g 44 g 36 g$

Thus, 16 g of methane reacts with 64 g of oxygen to produce 44 g of carbon dioxide.

The amount of oxygen required to react with 40 g of methane = $\frac{64}{16} \times 40 = 160 g$

Since only 48 g of oxygen is present, it will act as a limiting reagent.

The amount of carbon dioxide produced by 48 g of oxygen = $\frac{44}{64} \times 48 = 33 g$

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