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What quantity of copper (II) oxide will 2.80 litres of hydrogen at N.T.P. ? |
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Answer» `underset(underset("79.5 G 22.4 L at N.T.P")("1 mol1mol"))(CUO+H_(2))rarrCu+H_(2)O` 22.4 L of `H_(2)` REACT with CuO at N.T.P = 79.5 g 2.8 L of `H_(2)` react with CuO at N.T.P `= ((79.5g))/((22.4L))XX(2.8L)=9.94g`. |
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