1.

What quantity of copper(II) oxide will react 2.80litre of hydrogen at NTP

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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