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Q.

20 gms of sulphur on burning in air produced 11.2 its of SO2 at STP. The percentage of unreacted sulphur

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a

80%

b

20%

c

60%

d

40%

answer is B.

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Detailed Solution

S+O2SO2

1mol sulphate = 1 mole SO2=22.4 Lt SO2

11.2 Lt SO2=12mole sulphure=16 grams sulphur

% unreacted=420×100=20%

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