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) $\to$ 2H $_{2}$ O(l) + 3S(s)
Which of the following shows the volume of the gas remaining after the reaction?

0.2 litres of SO

_{2}

(g)

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0.4 litres of H

_{2}

(g)

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1.2 litres of H

_{2}

S(g)

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1.2 litres of SO

_{2}

(g)

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Solution

The correct option is A 0.2 litres of SO $_{2}$ (g)
2H $_{2}$ S(g) + SO $_{2}$ $\to$ 2H $_{2}$ O(l) + 3S(s)
2 1
$2.8 H_{2}$ S is limiting reagent
$S O_{2}$ remaining = $1.6 - 1.4 = 0.2 l i t$

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Similar questions

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?

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.

Calculate the volume of 320 g of SO₂ at STP.

[Atomic mass: S = 32 and 0=16]

Q. Volume (in litres) of 0.1M

H_{2} {S O}_{4}

required to completely neutralize 1 litres of 0.5M NaOH is:

5.6 litres of a gas weighs $1 g$ at STP. The gas can be:

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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 2H2s (g) + SO2(g) →2H2O(l) + 3S(s)Which of the following shows the volume of the gas remaining after the reaction?

The correct option is A 0.2 litres of SO2(g)2H2S(g) + SO2 → 2H2O(l) + 3S(s) 2 12.8 H2S is limiting reagent SO2 remaining = 1.6−1.4=0.2lit

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) $\to$ 2H $_{2}$ O(l) + 3S(s)
Which of the following shows the volume of the gas remaining after the reaction?

The correct option is A 0.2 litres of SO $_{2}$ (g)
2H $_{2}$ S(g) + SO $_{2}$ $\to$ 2H $_{2}$ O(l) + 3S(s)
2 1
$2.8 H_{2}$ S is limiting reagent
$S O_{2}$ remaining = $1.6 - 1.4 = 0.2 l i t$