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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» Step I. Calculation of mass of constituents in the mixture Mass of mixture = 1.0 g Let the mass of `Na_(2)CO_(3) = xg` Mass of `NaHCO_(3)=(1-x)g` Moles of `Na_(2)CO_(3)=("Mass of "Na_(2)CO_(3))/("Molar mass")=((xg))/(("106 g mol"^(-1)))=(x)/(106)` mol Moles of `NaHCO_(3)=("Mass of "NaHCO_(3))/("Molar mass")=((1-x)g)/(("84 g mol"^(-1)))=(1-x)/(84)`mol According to available data : Moles of `Na_(2)CO_(3)` = Moles of `NaHCO_(3)` `:. ((x)/(106)"mol")=((1-x)/(84)"mol")` or `84x=106-106xorx=(106)/(190)=0.558g` `:.` Mass of `Na_(2)CO_(3)` in the mixture = 0.558 g Mass of `NaHCO_(3)` in the mixture `= (1 - 0.558) =0.442 g` Step II. Calculation of total mass of `HCl` required `underset(106g)(Na_(2)CO_(3)(s))+underset(73g)(2HCl)(aq)rarr2NaCl(aq)+H_(2)O(l)+CO_(2)(g)` `underset(84g)(NaHCO_(3))(s)+underset(36.5g)(HCl(aq))rarrNaCl(aq)+H_(2)O(l)+CO_(2)(g)` Now, 106 g of `Na_(2)CO_(3)` require HCl = 73 g 0.558 g of `Na_(2)CO_(3)` require `HCl=((73g)xx(0.558g))/((106g))=0.384g` Similarly, 84 g of `NaHCO_(3)` require HCl = 36.5 0.442 g of `NaHCO_(3)` require HCl `= ((36.5g)xx(0.442g))/((84g))=0.192g` `:.` Total mass of HCl required `= (0.384 + 0.192) = 0.576 g` Step III. Calculation of volume of `HCl` required Mass of HCl required = 0.576 g Molarity of HCl solution = 0.1 M Molarity of solution (M) `= ("Mass of HCl/Molar mass of HCl")/("Volume of HCl solution (V)")` `("0.1 mol L"^(-1))=((0.576g)//(36.5g//mol))/(V)` `= V = ((0.576g))/(("36.5 gmol"^(-1))xx("0.1 mol L"^(-1)))=0.1578L =157.8mL`. |
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