In our problem both
Na2Co3 and NaHCO3 both are equimolar that means molairy of both substance are
equal
So number of moles of
Na2CO3 = number of moles of NaHCO3 --------------(1)
Let find moles of both
substances
Let take mass of Na2CO3
= x g
Then mass of NaHCO3 = 1-x g
Molar mass of Na2CO3 =
2*23+ 12+ 3*16 = 106 g/ mol
Molar mass of NaHCO3 =
1*23 + 1 + 12 + 3*16 = 84 g/mol
So number of moles of
Na2CO3 = mass/ molar mass = x/106 moles ……….(2)
Number of moles of
NaHCO3 is = (1-x)/ 84 gram
Plug the value in
equation (1)
We get
x/106 = 1-x/84
cross multiply them we
get
84x = 106 - 106 x
solve it we get
X = 106/190 = 0.558 gram
Plug the value back in
equation (2) we get = 0.558/106 =
0.00526 moles
Our reactions will
Na2CO3 + 2HCl ----> 2NaCl +
H2CO3
NaHCO3+ HCl ------> NaCl + H2CO3
Here in our reaction 2
moles of HCl is required each for of
Na2CO3 so that
Moles of HCl required to react with Na2CO3 =
2* 0.00526 = 0.01052
And moles of HCl
required for NaHCO3 will also 0.00526
Total number of HCl
moles required = 0.01052 + 0.00525 = 0.01578 moles
Molarity of HCl = 0.1M
given in our problem
Molarity = number of
moles of solute / volume of solution in liter
0.1 =
0.01578/ volume in liter
Cross multiply now we
get
0.1 *
volume in liter = 0.01578
Divide by 0.1 we get
Volume in liter = 0.01578/0.1 = 0.1578 liter
Volume in ml = 0.01578
liter * 1000 ml / 1liter = 157.8 ml
