${CH}_{4} (g) + 2 O_{2} (g) ⟶ {CO}_{2} (g) + 2 H_{2} O (g)$
According to the above chemical reaction, following information have been given.
(i) One mole of ${CH}_{4}$ (g ) reacts with two moles of $O_{2}$ (g) to give one mole of ${CO}_{2} (g)$ and two moles of $H_{2} O$ (g)
(ii) One molecule of ${CH}_{4}$ (g) reacts with 2 molecules of $O_{2} (g)$ to give one molecule of ${CO}_{2} (g)$ and 2 molecules of $H_{2} O (g)$
(iii) 22.4 L of ${CH}_{4}$ (g ) reacts with 45.4 L of $O_{2}$ to give 22.7 L of ${CO}_{2} (g)$ and 45.4 L of $H_{2} O (g)$
(iv) 16 g of ${CH}_{4}$ (g ) reacts with 2x 32 g of $O_{2}$ (g ) to give 44 g of ${CO}_{2} (g)$ and 2x 18 g of $H_{2} O (g)$
From these information the true interconversion should be

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mass = moles = number of molecules

mass $\overset{}{⇌}$ moles $\overset{}{⇌}$ number of molecules

rnass $\to$ moles $\overset{}{⇌}$ nurnber of molecules

rnass $\to$ moles $\to$ number of molecules

answer is B.

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Detailed Solution

From the given relationships, the data can be interconverted as

mass $\overset{}{⇌}$ moles $\overset{}{⇌}$ number of molecules

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