2.1 g of a mixture of `NaHCO_3` and `KCIO_3` required 100 " mL of " 0.1 HCl for complete reaction. Calculate the amount of residue that would be obtained on heating 2.2 g of the same mixture strongly.

Explanation

`Ew(NaHCO_3)=23+1+12+3xx16=(84)/(1)=84(n=1)`
HCl reacts only with `NaHCO_3` (acid and base reaction)
`m" Eq of "HCl=100xx0.1`
`=10m" Eq of "NaHCO_3`
`=10xx10^(-3)xx84g of NaHCO_3=0.84g`
Weight of `KClO_3=2.1-0.84=1.26g`
Weight of `NaHCO_3` in `2.2 g of mixture=(0.84xx2.2)/(2.1)`
`=0.88g`
Weight of `KClO_3` in 2.2 g of mixture `=(2.2-0.88)g`
`=1.32`g
Heating of mixture:
`NaHCO_3overset(triangle)toNa_2CO_3+H_2O+CO_2`
Weight of residue obtained on heating `NaHCO_3`
`=(106xx0.88)/(2xx84)=0.555g`
`2KClO_3to2KCl+3O_2`
Weight of residue obtained on heating `KClO_3`
`=(2xx74.5xx1.32)/(2xx122.5)=0.802g`
Total weight of residue `=0.555+0.802=1.357g`