The correct option is C 156 mL
Na2CO3nf=2+2HCl→2NaCl+Na2CO3
NaHCO3nf=1+HCl→NaCl+CO2+H2O
Let x g of Na2CO3 and (1−x) g of NaHCO3 are present in the mixture.
Equivalents of Na2CO3+ Equivalents of NaHCO3= Equivalents of HCl
∴ x106×2+1−x86×1=0.1×V ...(1)
Given, Moles of Na2CO3=Moles of NaHCO3
∴ x106=1−x86 ...(2)
86x=106−106x
∴ x=0.552
From equation (1),
0.552106×2+1−0.55286×1=0.1×V
0.0104+0.0052=0.1×V
0.0156=0.1×V
∴ V=0.156 L
or, V=156 mL