Mass of $O_{2}, N_{2}$ and $C H_{4}$ containing equal moles will be in the ratio of :

16 : 14 : 16

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8 : 7 : 4

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1 : 1 : 1

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4 : 7 : 4

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Solution

The correct option is C $8 : 7 : 4$
Ratio of one mole of $O_{2}, N_{2}$ and $C H_{4}$ is $32 : 28 : 16 = 8 : 7 : 4$ .

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Similar questions

1, 2, 4, 4, 7, 8, ?, 16

(\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \frac{1}{10 \times 13} + \frac{1}{13 \times 16})

$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \frac{1}{10 \times 13} + \frac{1}{13 \times 16} + \frac{1}{16 \times 19}$ =

N_{2}

O_{2}

4 : 1

4 $^{7}$ $\div$ 16 $^{4}$ x $\sqrt{16}$ = ?

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