The amount of water produced (in $g$ ) in the oxidation of 1 mole of rhombic sulphur by conc. $H N O_{3}$ to a compound with the highest oxidation state of sulphur is
(Given data: Molar mass of water = 18 $g m o l^{- 1}$ )

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Solution

The corresponding reaction of rhombic sulphur and conc. $H N O_{3}$ is
$S_{8} + 48 H N O_{3} \to 8 H_{2} S O_{4} + 48 N O_{2} + 16 H_{2} O$

Mass of $H_{2} O = 18 \times 16 = 288 g$

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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 gmol−1)

The corresponding reaction of rhombic sulphur and conc. HNO3 is S8+48HNO3→8H2SO4+48NO2+16H2O Mass of H2O=18×16=288 g

The amount of water produced (in $g$ ) in the oxidation of 1 mole of rhombic sulphur by conc. $H N O_{3}$ to a compound with the highest oxidation state of sulphur is
(Given data: Molar mass of water = 18 $g m o l^{- 1}$ )

The corresponding reaction of rhombic sulphur and conc. $H N O_{3}$ is
$S_{8} + 48 H N O_{3} \to 8 H_{2} S O_{4} + 48 N O_{2} + 16 H_{2} O$

Mass of $H_{2} O = 18 \times 16 = 288 g$