2 mol of `H_(2)S` and 11.2 L of `SO_(2)` at N.T.P. react to form x moles of sulphur, x is `SO_(2)+2H_(2)S to 3S +2H_(2)O`

Question

2 mol of `H_(2)S` and 11.2 L of `SO_(2)` at N.T.P. react to form x moles of sulphur, x is `SO_(2)+2H_(2)S to 3S +2H_(2)O`

2 mol of `H_(2)S` and 11.2 L of `SO_(2)` at N.T.P. react to form x moles of sulphur, x is
`SO_(2)+2H_(2)S to 3S +2H_(2)O`
A. `1.5`
B. `3`
C. `11.2`
D. `6`

Monjur Alahi

4 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 - A
22.4 L of `SO_(2)` at N.T.P. = 1 mol
11.2 L of `SO_(2)` at N.T.P. = 0.5 mol
`underset(1mol)(SO_(2))+underset(2mol)(2H_(2)S)to3S+2H_(2)O`
Here `SO_(2)` is the limiting reactant
`:.` Moles of sulphur formed `= 3xx"moles of" SO_(2)`
`= 3xx0.5mol`
`= 1.5 mol`

4 views Answered 2023-02-05 15:29