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What quantity of copper(II) oxide will react 2.80litre of hydrogen at NTP |
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Answer» The balanced equation is `underset("79.5g")underset(1"mol")(CuO+H_(2)) to underset(22.4"litre at NTP")underset("1mol")(Cu)+H_(2)O` 22.4 litre of hydrogen at NTP reduce CuO=79.5g 2.80 litre of hydrogen at NTP will reduce CuO `=(79.5)/(22.4)xx2.80g=9.95g` |
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