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`

Options

A. `1.5`
B. `3`
C. `11.2`
D. `6`

Answer

A

Explanation

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`