let the amount of Na₂CO₃ be x

& that of NaHCO₃ be 1-x

Now moles of Na₂CO₃ = x / 106

& moles of NaHCO₃ = 1-x / 84

Now according to question , number of moles of Na₂Co₃ = number of moles of NaHCO₃

Therefore x / 106 = 1-x / 84

84x = 106-106x

84x +106x = 106

190x = 106

Or

x = 106 / 190 = 0.558

Therefore moles of Na₂Co₃ = 0.558 / 106 = 0.00526

&

moles of NaHCO₃ = 1 - 0.558 / 84 = 0.0053

Now Hcl reacts with Na₂Co₃ & NaHCO₃ as follows:

Na₂Co₃ + 2Hcl  2Nacl + H₂o + CO₂

NaHCO₃ + Hcl  Nacl + H₂o + CO₂

From the above reactions, 1 mol of Na₂Co₃ will react with 2 mol of Hcl

Therefore 0.00526 mol of Na2Co₃ will react with 2 x 0.00526 mol of Hcl & similarly 0.00526 mol of NaHCO₃ will react with 0.00526 mol of Hcl

Total moles of Hcl required to react with mixture of of NaHCO₃ & Na₂Co₃

= 2 X 0.00526 + 0.00526 =0.01578 mol

Also according to question 0.1 mol of 0.1 M Hcl is present in 1000 ml

Or

0.01578 mol of 0.1 M Hcl is present in (1000/0.1) x 0.01578 = 158 ml

Hence, 158 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na₂CO₃ and NaHCO_3, containing equimolar amounts of both.