Volume of `0.1M HCl` required to react completely with 1g equimolar mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` is

Explanation

Calculate of no. of moles of components in the mixture
Let x g of `Na_(2)CO_(3)` is present in the mixture.
`therefore(1-x)g` of `NaHCO_(3)` is present in the mixture.
Molar mass of `Na_(2)CO_(3)` ,brgt `=2xx23+12+3xx16=106" g "mol^(-1)`
and molar mass of `NaHCO_(3)`
`=23xx1+1+12+3xx16=84" g "mol^(-1)`
No. of moles of `Na_(2)CO_(3)` in x g`=(x)/(106)`
No. of moles of `NaHCO_(3)` in (1-x)`g=(1-x)//84`
As given that the mixture contains equimolar
amounts of `Na_(2)CO_(3)` and `NaHCO_(3)` therefore.
`(x)/(106)=(1-x)/(84)`
`=106-106x=84x`
`106=190x`
`thereforex(106)/(190)=0*558g`
`therefore` No. of moles of `Na_(2)CO_(3)` present
`=(0*558)/(106)=0*00526`
and no. of moles of `NaHCO_(3)` present
`=(1-0*558)/(84)=0*00526`
Calculation of no. of moles of of HCl required
`Na_(2)CO_(3)+2HClto2NaCl+H_(2)O+CO_(2)`
`NaHCO_(3)+HClto NaCl+H_(2)O+CO_(2)`
As can be seen, each mole of `Na_(2)CO_(3)` needs ltBrgt 2 moles of HCl,
`therefore0*00526` mole of `Na_(2)CO_(3)` needs
`=0*00526xx2=0*01052` moles
Each mole of `NaHCO_(3)` needs 1 mole of HCl.
`0*00526=0*00526` mole
Total amount of HCl needed will be `=0*01052+0*00526=0*01578` mole
`0*1` mole of 0.1 M HCl are present in 1000 mL of HCl
`therefore0*01578` mole of `0*1` M HCl will be present n
`=(1000)/(0*1)xx0*01578=157*8mL`