10 grams of hydrogen and 142 grams of chlorine are made to react with each other. The remaining amount of reactant gas is made to react with excess oxygen. Same amount of the above reactant gas is made to react with nitrogen. The amounts of the products formed in both reactions.

water - 54gm ; ammonia - 34gm

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water - 34gm ; ammonia - 54gm

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water - 17gm ; ammonia - 34gm

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water - 27gm ; ammonia - 14gm

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Solution

The correct option is A water - 54gm ; ammonia - 34gm
$H_{2} + C l_{2} ⟶ 2 H C l$
$10 g o f h y d r o g e n = 5 m o l e s$
$142 g o f c h l o r i n e = 2 m o l e s$
2 moles chlorine reacts with 2 moles hydrogen to form 2 moles $H C l$ , 3 moles of hydrogen remains unreacted.
$2 H_{2} + O_{2} ⟶ 2 H_{2} O$
$2 m o l e s o f h y d r o g e n ⟶ 2 m o l e s o f w a t e r$
$3 m o l e s o f h y d r o g e n ⟶ 3 m o l e s o f w a t e r$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$3 m o l e s o f h y d r o g e n ⟶ 2 m o l e s o f N H_{3}$
$3 m o l e s o f w a t e r = 54 g$
$2 m o l e s o f N H_{3} = 34 g$

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