What weight of zinc and sulphuric acid will be required to produce enough hydrogen to reduce completely 8.5g of CuO to Cu? (Zn=65.3, Cu=63.5)

Explanation

The reactions involved are- < br> (i). `Zn+H_(2)SO_(4) to ZnSO_(4)+H_(2)`
(ii).`CuO+H_(2) to Cu+H_(2)O`
Thus, `(63.5+16)=79.5g` of CuO reacts reacts with 2g of hydrogen.
`therefore` 8.5g of CuO reacts with `(2times8.5)/(79.5)=0.2138`g of hydrogen
Again 2.0g of hydrogen is produced from 63.5g of Zn.
`therefore` 0.2138g of hydrogen is produced from `(65.3times0.2138)/(2)g=6.98g` of Zn
Now, `underset(65.3g)(Zn)+underset((2+32+16times4)=98g) (H_(2)SO_(4)) to ZnSO_(4)+underset(4g)(H_(2))`
Thus, 2g of hydrogen is produced by 98g of `H_(2)SO_(4)`
`therefore` 0.2138g of hydrogen is produced by `(98times0.2138)/(2)g=10.47g` of `H_(2)SO_(4)`
Hence, 6.98g of zinc and 10.47g of `H_(2)SO_(4)` will be required to produce enough hydrogen to completely reduce 8.5g of CuO and Cu.
All these problems can be solved in terms of numbers of mole as well.