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Standard XII

Chemistry

Faraday's First Law

10 L of N2 ...

Question

10 L of $N_{2}$ and 30 L of $H_{2}$ combine to form $N H_{3}$ having volume of:

20 L

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40 L

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30 L

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10 L

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Solution

The correct option is A 20 L
$N_{2} + 3 H_{2} \to 2 N H_{3}$

Here, 1 L $N_{2}$ combines with 3 L of $H_{2}$ to form 2 L of $N H_{3}$ .

Therefore, volume of ammonia $= 2 \times$ volume of $N_{2}$ $= 2 \times 10 = 20$ L

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Similar questions

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Find out range and coefficient of range of the following series.

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10 L of N2 and 30 L of H2 combine to form NH3 having volume of:

The correct option is A 20 L N2+3H2→2NH3Here, 1 L N2 combines with 3 L of H2 to form 2 L of NH3.Therefore, volume of ammonia =2× volume of N2 =2×10=20 L

10 L of $N_{2}$ and 30 L of $H_{2}$ combine to form $N H_{3}$ having volume of:

The correct option is A 20 L
$N_{2} + 3 H_{2} \to 2 N H_{3}$

Here, 1 L $N_{2}$ combines with 3 L of $H_{2}$ to form 2 L of $N H_{3}$ .

Therefore, volume of ammonia $= 2 \times$ volume of $N_{2}$ $= 2 \times 10 = 20$ L