Arun
Last Activity: 5 Years ago
Dear student
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
Regards
Arun (askIITians forum expert)
