1 mol of SO2 and 1 mol of H2S react completely to form H2O and S as follows :SO2 + 2H2S arrow 2H2O + 3 S (At. mass S = 32,O = 16)The mass of S obtained is;

Question

1 mol of SO2 and 1 mol of H2S react completely to form H2O and S as follows :SO2 + 2H2S arrow 2H2O + 3 S (At. mass S = 32,O = 16)The mass of S obtained is; (A) 96 g S (B) 48 g S (C) 24 g S (D) 64 g S

Tardigrade Admin · Accepted Answer

SO2+H2S arrow2H2O+3S Here H2S is the limiting reactant 2 mol of H2S give 3 mol of S ∴ 1 mol of H2S gives 3/2 mol of S Mass of sulphur =((3/2)mol)×(32 g mol-1) =48 g

Q. 1 mol of $S O_{2}$ and 1 mol of $H_{2} S$ react completely to form $H_{2} O$ and S as follows : $S O_{2} + 2 H_{2} S \to 2 H_{2} O + 3 S$ (At. mass S = 32,O = 16)The mass of S obtained is;

96 g S

48 g S

24 g S

64 g S

$S O_{2} + H_{2} S \to 2 H_{2} O + 3 S$
Here $H_{2} S$ is the limiting reactant
$2$ mol of $H_{2} S$ give $3$ mol of $S$
$∴$ $1$ mol of $H_{2} S$ gives $3/2$ mol of $S$
Mass of sulphur $= (\frac{3}{2} m o l) \times (32 g m o l^{- 1})$
$= 48 g$

Q. 1 mol of SO2​ and 1 mol of H2​S react completely to form H2​O and S as follows :SO2​+2H2​S→2H2​O+3S (At. mass S = 32,O = 16)The mass of S obtained is;