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प्रश्न
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate
- the volume of gaseous product formed.
- composition of the resulting mixture.
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उत्तर
\[\ce{H2S + Cl2 -> 2HCl + S}\]
(i)
At STP,
1 mole gas occupies = 22.4 L.
1 mole H2S gas produces = 2 moles HCl gas,
∴ 22.4 L H2S gas produces
= 22.4 × 2
= 44.8 L HCl gas.
Hence, 112 cm3 H2S gas will produce
= 112 × 2
= 224 cm3 HCl gas.
Hence, 224 cm3 HCl gas is produced.
(ii)
1 mole H2S gas consumes = 1 mole Cl2 gas.
Hence, 22.4 L H2S gas consumes = 22.4 L Cl2 gas at STP.
∴ 112 cm3 H2S gas consumes = 112 cm3 Cl2 gas.
120 cm3 − 112 cm3 = 8 cm3 Cl2 gas remains unreacted.
Hence, the composition of the resulting mixture is 224 cm3 HCl gas + 8 cm3 Cl2 gas.
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