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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` |
Answer» 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` |
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