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

Only `NaHCO_(3)` reacts with 0.1 N HCl let x g be the weight of `NaHCO_(3)` and `(2.1-x)g ` be the weight of `KClO_(3)`
m" Eq of "`NaHCO_(3)=100xx0.1=10`
Weight of `NaHCO_(3)=10xx10^(-3)xx84=0.84g`
Weight of `KClO_(3)=2.1-0.84=1.26g`
`2NaHCO_(3)overset(Delta)toNa_(2)CO_(3)+H_(2)Ouarr+CO_(2)uarr`
`{:([ 2 " mol of "NaHCO_(3)=2xx82g),(Na_(2)CO_(3)=106g)]:}`
`2KClO_(3)overset(Delta)to2KCl+3O_(2)uarr`
The residue will have `Na_(2)CO_(3)` and `KCl` `{:([KClO_(3)=122.5g),(KCl=74.5g)]:}`
Weight of redidue from 2.1 g of mixture
`=(53)/(84)xx0.84+(74.5)/(122.5)xx1.26`
`=053+0.766=1.296g`
Weight of residue from 2.2 g of the mixture
`=0.84+1.26=2.20g`
`=(1.296)/(2.1)xx2.2=1.358g`