Step 1 : Calculation of the number of moles of the components in the mixture.
Let Na2 CO3 present in the mixture =x g
∴‌‌NaHCO3 present in the mixture =(1−x) g
Molar mass of Na2 CO3 =2×23+12+3×16 =106 g ‌mol−1
Molar mass of NaHCO3 =23+1+12+3×16 =84 g mol−1
∴‌ No. of moles of Na2CO3 in x g=

x

106

No. of moles of NaHCO3 in(1−x)g=

1−x

84

As mixture contains equimolar amounts of the two,

x

106

=

1−x

84

or, 106−106x=84x or x=

106

190

g=0.558g
Thus, moles of Na2CO3=

0.558

106

=0.00526
∴‌ Moles of NaHCO3=0.00526
Step 2 : To calculate the moles of HCl required.
Na2CO3+2‌HCl→2‌NaCl+H2O+CO2
NaHCO3+HCl→NaCl+ H2O+CO2
1 mole of Na2CO3 requires ‌HCl=2 moles
∴‌0.00526 mole of Na2CO3 requires ‌HCl=0.00526×2=0.01052 ‌mole
1 mole of NaHCO3 requires HCl=1 ‌mole
∴‌0.00526 mole of NaHCO3 requires ‌HCl=0.00526‌mole
∴‌ Total ‌HCl required =0.01052+0.00526=0.01578‌mole
Step 3 : To calculate volume of 0.1 M HCl
Volume of acid =

Number‌of‌moles

Molarity

=

0.01578

0.1

=157.8 ‌mL