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?
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 = 157.8 ml
Hence, 158 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.
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