Question

How many `mL` of a `0.1M HCl` are required to react completely with `1 g` mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of two?

Answer 1

Calculation of mass of constitunets in the mizture.
Mass of the mixture=1.0g
Let the mass of `Na_(2)CO_(3)` =x g
Mass of `NaHCO_(3)`=(1-x)g
`"Moles of"NaHCO_(3)=("Mass of" Na_(2)CO_(3))/("Molar mass")=((xg))/((106g mol^(-1)))=x/106 mol`
`"Moles of" NaHCO_(3)=("Mass of "NaHCO_(3))/("Molar mass")=((1-x)g)/((84gmol^(-1)))=(1-x)/84mol`
According to available date:
`"Moles of"Na_(2)CO_(2)="Moles of "NaHCO_(3`)
`(x/106 mol)=((1-x)/84mol)`
84x=106-106x or x=0.442 g
Mass of `Na_(2)CO_(3)` in the mixture=(0.558 g
Mass of `NaHCO_(3)` in the mizture=(1-0.558)=0.442 g
Calculation of total mass of HCI required. `Na_(2)CO_(3)(s)+2HCI(aq)to2NACI(aq)+H_(2)O(I)+CO_(2)(g)`
`underset(106g)(Na_(2)CO_(3)(s))+underset(73g)(2HCl(aq)) rarr 2Na(aq)+H_(2)O(l)+CO_(2)(g)`
`underset(84g)(NaHCO_(3)(s))+underset(36.5g)(HCI(aq))to2NaCI("aq")+H_(2)O(I)+CO_(2)(g)`
106 g of `Na_(2)CO_(3)` require HCI=73g
0.55g of `Na_(2)CO_(3)` require HCI=`((73g)xx(0.558g))/((106g))=0.384 g`
Similarly, 84 g of require HVI=36.5 g
0.442 g of `NaHCO_(3)` require HCI=`((36.5g)xx(0.442g))/((84g))=0.192`
Total mass of HCI required = (0,384+0.192)=0.576 g
Calculation of volum of HCI required.
Mass of HCI required = 0.567 g
Molarity of HCI solution = 0.1 M
`"Molarity of solution (M)"= ("Mass of HCI / molar mass of HCI")/("Volume of solution (V)")`
`0.1 mol^(-1)=((0.576 g)xx(36.5g//mol))/V`
`V=(0.576xx1000)/(36.5xx0.1) =0.1578 L=157.8 mL`