S+O2→SO2 (Yield = 80%)SO2+Cl2+2H2O→H2SO4+2HCl (Yield = 60%)H2SO4+ – InterviewSolution

InterviewSolution

1.	S+O2→SO2 (Yield = 80%)SO2+Cl2+2H2O→H2SO4+2HCl (Yield = 60%)H2SO4+BaCl2→BaSO4+2HCl (Yield = 30%)8 g of Sulphur is burnt to form SO2, which is oxidized by Cl2 water. The solution is then treated with BaCl2 solution. The amount of BaSO4 precipitated is:(Molar mass of Barium = 137 g/mol)
Answer» $S + O_{2} \to S O_{2} (Yield = 80%)$ $S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)$ $H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)$ 8 g of Sulphur is burnt to form $S O_{2}$ , which is oxidized by $C l_{2}$ water. The solution is then treated with $B a C l_{2}$ solution. The amount of $B a S O_{4}$ precipitated is: (Molar mass of Barium = $137 g/mol$ )

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