Let the mixture contains x g of sodium carbonate and 1 - x g of sodium bicarbonate.

The molar masses of sodium carbonate and sodium bicarbonate are 106 g/mol and 84 g/mol respectively.

The number of moles of sodium carbonate and sodium bicarbonates are x/106 and 1 − x/84 respectively.

Since, it is an equilmolar mixture, x/106 = 1 − x/84(84x) = 106 − 106 x 190 x = 106x = 0.5579

Number of moles of sodium carbonate = 0.5579/106 = 0.005263

Number of moles of sodium hydrogen carbonate = 1 - 0.5579 84 = 0.005263

One mole of sodium carbonate will react with 2 moles of HCl and 1 mole of sodium bicarbonate will react with 1 mole of HCl.

Total number of moles of HCl that will completely neutralize the mixture = 2 × 0.005263 + 0.005263 = 0.01578 moles

Volume of 0.1 M HCl required = 0.01578/0.1 = 0.158 L = 158 mL.