1 mol of `SO_(2)` and 1 mol of `H_(2)S` react completely to form `H_(2)O` and S as follows: `SO_(2)+2H_(2)S to 2H_(2)O+3S` (At. mass S = 32, O = 16) T

Question

1 mol of `SO_(2)` and 1 mol of `H_(2)S` react completely to form `H_(2)O` and S as follows: `SO_(2)+2H_(2)S to 2H_(2)O+3S` (At. mass S = 32, O = 16) T

1 mol of `SO_(2)` and 1 mol of `H_(2)S` react completely to form `H_(2)O` and S as follows:
`SO_(2)+2H_(2)S to 2H_(2)O+3S`
(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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`SO_(2)+H_(2)Sto2H_(2)O+3S`
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 `= ((3)/(2)mol)xx(32gmol^(-1))`
`=48 g`

5 views Answered 2023-02-05 15:29