How many of 0.1 N HCl are required to react completely with 1 g mixture of `Na_2CO_3 and NaHCO_3` containing equimolar amounts of two ?

Explanation

Step 1. To calculate the number of moles of the components in the mixture.
Suppose `Na_(2)CO_(3)` present in the mixture = xg`" "therefore" "NaHCO_(3)" present in the mixture "=(1-x)g`
`"Molar mass of "Na_(2)CO_(3)=2xx23+12+3xx16="106 g mol"^(-1)`
`"Molar mass of "NaHCO_(3)=23+1+12+3xx16="84 g mol"^(-1)`
`therefore" Moles of "Na_(2)CO_(3)" in x g "=(x)/(106)," Moles of "NaHCO_(3)" in "(1-x)g=(1-x)/(84)`
As mixture contains equimolar amounts of the two.
`(x)/(106)=(1-x)/(84)" or "106-106x=84x or x=(106)/(190)g=0.558g`
Thus,`" moles of "Na_(2)CO_(3)=(0.558)/(106)=0.00526," Moles of "NaHCO_(3)=(1-0.558)/(84)=(0.442)/(84)=0.00526`
Step 2. To calculate the moles of HCl required.
`Na_(2)CO_(3)+2HCl rarr 2NaCl+H_(2)O+CO_(2),NaHCO_(3)+HCl rarr NaCl +H_(2)O+CO_(2)`
`"1 mole of "Na_(2)CO_(3)" required HCl = 2 moles"`
`therefore" 0.00526 mole of "Na_(2)CO_(3)" requires HCl "=0.00526 xx"2 moles = 0.01052 mole"`
`" 1 mole of "NaHCO_(3)" requires HCl = 1 mole"`
`therefore" 0.00526 mole of "NaHCO_(3)" required HCl"="0.00526 mole"`
`therefore" Total HCl required "=0.01052+0.00526" moles = 0.01578 moles"`
Step 3. To calculate volume of 0.1 M HCl
`"0.1 mole of 0.1 M HCl are present in 1000 mL of HCl."`
`"0.01578 mole of 0.1 M HCl will be present in HCl "=(1000)/(0.1)xx0.01578="157.8 mL"`