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Standard XII

Chemistry

Empirical & Molecular Formula

At NTP, 10 li...

Question

At NTP, 10 litre of hydrogen sulphide gas reacted with 10 litre of sulphur dioxide gas. The volume of gas, after the reaction is complete, would be:

5 litre

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10 litre

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15 litre

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20 litre

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Solution

The correct option is B 5 litre

$2 H_{2} S + S O 2 \to 2 H_{2} O + 3 S$
According to the equation $2 \times 22.4$ litre of $H_{2} S$ gas reacts with $22.4$ litre of $S O_{2}$
So $5$ litre of $H_{2} S$ reacts with $10$ litre of $S O_{2}$
Hence option $A$ is correct.

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Similar questions

Q. At NTP,

10

liter of hydrogen sulphide gas reacted with

10

liter of suphur dioxide gas. The volume of gas, after the reaction is completed would be:

Q. At STP,

2.8

litres of hydrogen sulphide were mixed with

1.6

litre of sulphur dioxide and the reaction occurred according to the equation:

2 H_{2} S (g) + S O_{2} (g) \to 2 H_{2} O (l) + 3 S (s)

Which of the following shows the volume of the gas remaining after the reaction?

Q. 10 litre of

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Q. At STP , 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation 2H

_{2}

s (g) + SO

_{2}

(g)

\to

_{2}

O(l) + 3S(s)
Which of the following shows the volume of the gas remaining after the reaction?

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:
$2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$
(i). calculate the volume of hydrogen suulphide at STP.
Also, (ii). calculate the volume of oxygen required at STP. Which will complete the combustion of hydrogen sulphide determined in litres.

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At NTP, 10 litre of hydrogen sulphide gas reacted with 10 litre of sulphur dioxide gas. The volume of gas, after the reaction is complete, would be:

The correct option is B 5 litre 2H2S+SO2→2H2O+3S According to the equation 2×22.4 litre of H2S gas reacts with 22.4 litre of SO2So 5 litre of H2S reacts with 10 litre of SO2Hence option A is correct.

The correct option is B 5 litre

$2 H_{2} S + S O 2 \to 2 H_{2} O + 3 S$
According to the equation $2 \times 22.4$ litre of $H_{2} S$ gas reacts with $22.4$ litre of $S O_{2}$
So $5$ litre of $H_{2} S$ reacts with $10$ litre of $S O_{2}$
Hence option $A$ is correct.