How many ml of 0.1 m hcl are required


Asked by admin @ in Chemistry viewed by 290 People


How many ml of 0.1m hcl is required to react completely with 1g mixture of na2co3 and nahco3 containing equal-molar amounts of both?

Answered by admin @



Na2Co3 and NaHCO3 are both equimolar. It is the meaning of substances to have the same molarity

Na2CO3 in the mixture with a total amount will be x gram

NaHCO3 in the mixture with a total amount will be (1-x) gram

Now the molar mass of Na2CO3 = 106g/mol

therefore,

So the nuber of moles is Na2CO3 = x/106mol

molar mass of NaHCO3 = 84 g/mol

Therefore, number of moles of NaHCO3 = 1-x/84 mol

As per the given questions

X/106 = 1-X/54

84x = 106 – 106x

190x = 106

x = 0.5579

After that

number of moles of Na2CO3 = 0.5579/106 mol

                                             = 0.0053mol

And number of moles of NaHCO3 = 1 – 0.5579/84

                                                   = 0.0053 mol

Hcl reacts with Na2CO3 and Na2CO3 according to the following equation

2HCl + Na2CO3 ------------> 2NaCl + H2O + CO2

2mol      1mol                  

and

HCl + NaHCO3 ----------->NaCl  + H2O + CO2

1mol     1 mol

From the above reactions,

1 mol of Na2Co3 will react with 2 mol of Hcl

Therefore, 0.00526 mol of Na2Co3 will react with 2x 0.00526 mol of Hcl

similarly 0.00526 mol of NaHCO3 will react with 0.00526 mol of Hcl

Total mol of Hcl required to react with a mixture of NaHCO3 & Na2Co3= 2x 0.00526 + 0.00526= 0.01578 mol.


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