Question

M of HCl is 0.1,find volume of HCl required to completely react with 1g mixture of Na_{​​​​​2}​​​​CO_​​3 and NaHCO_​​3 containing equimolars amount of the two

Answer 1

Let the amount of Na2CO3 in the mixture will be x gram
Then,amount of NaHCO3 in the mixture will be (1-x) gram
molar mass of Na2CO3 = 106g/mol
therefore,
number of moles of Na2CO3 = x/106mol
molar mass of NaHCO3 = 84 g/mol
therefore,
number of moles of NaHCO3 = 1-x/84 mol
ACCORDING TO THE QUESTION,

X/106 = 1-X/54
84x = 106 – 106x
190x = 106
x = 0.5579
therefore,
number of moles of Na2CO3 = 0.5579/106 mol
= 0.0053mol
And number of moles of NaHCO3 = 1 – 0.5579/84
= 0.0053 mol
Hcl reacts with Na2CO3 and Na2CO3 according to the following equation


2HCl + Na2CO3 ------------> 2NaCl + H2O + CO2
2mol 1mol
and
HCl + NaHCO3 ----------->NaCl + H2O + CO2
1mol 1 mol
so,
1 mol of Na2CO3 reacts with 2 mol of HCl
therefore,0.0053 mol of Na2CO3 reacts with 2X0.0053 mol = 0.0106mol
and,
1 mol of NaHCO3 reacts with 1 mol of HCl
therefore,0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl
total moles of HCl = 0.0106 + 0.0053
= 0.0159 moles
In 0.1 mol HCL
0.1 mol of HCl is present in 1000ml of solution
therefore,
0.0159 mol of HCL is present in
1000 X 0.0159/0.1
= 159 ml of solution
Hence 159 ml of HCl is required to react completely with 1g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both