How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

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