6.4 g of sulphur is heated with an excess of hot concentrated sulphuric acid according to S + 2H2SO4 -> 3SO2 + 2H2O. Volume of SO2 liberated at STP is (S = 32; molar volume 22.4 L/mol)

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JEE Advanced levelSingle correct optionDifficulty 3 of 3pn159

$6.4 g$ of sulphur is heated with an excess of hot concentrated sulphuric acid according to $S + 2 H X_{2} SO X_{4} 3 SO X_{2} + 2 H X_{2} O$ . Volume of $SO X_{2}$ liberated at STP is (S = 32; molar volume 22.4 L/mol)

A $13.44 L$
B $8.96 L$
C $6.72 L$
D $4.48 L$

Answer: A

n(S) = 6.4/32 = 0.20 mol. The equation gives 3 mol SO2 per mol S, so n(SO2) = 0.60 mol. Volume = 0.60 x 22.4 = 13.44 L. Here S is oxidised (0 to +4) and H2SO4 is reduced (+6 to +4).

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6.4 g of sulphur is heated with an excess of hot concentrated sulphuric acid according to S+2HX2​SOX4​​3SOX2​+2HX2​O. Volume of SOX2​ liberated at STP is (S = 32; molar volume 22.4 L/mol)

Answer: A

More questions on Sulphur allotropes, SO2, sulphuric acid, oxyacids of sulphur

$6.4 g$ of sulphur is heated with an excess of hot concentrated sulphuric acid according to $S + 2 H X_{2} SO X_{4} 3 SO X_{2} + 2 H X_{2} O$ . Volume of $SO X_{2}$ liberated at STP is (S = 32; molar volume 22.4 L/mol)