JEE Advance - Chemistry (2019 - Paper 2 Offline - No. 13)
The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3 to a compound with the highest oxidation state of sulphur is ..............
(Given data : Molar mass of water = 18 g mol$$-$$1)
(Given data : Molar mass of water = 18 g mol$$-$$1)
Answer
288
Explanation
When rhombic sulphur (S8) is oxidised by conc. HNO3 then H2SO4 is obtained and NO2 gas is released.
$${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$$
1 mole of rhombic sulphur produces = 16 moles of H2O
$$ \therefore $$ Mass of water = 16 $$ \times $$ 18 (molar mass of H2O) = 288 g
$${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$$
1 mole of rhombic sulphur produces = 16 moles of H2O
$$ \therefore $$ Mass of water = 16 $$ \times $$ 18 (molar mass of H2O) = 288 g
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