Question

how many Ml of 0.1 M hcl are required to react completely with 1g mixture of Na2co3 and Na2co3 containing equimolar amount of both?? ​

Answer 1

Let the amount of Na2CO3 in the mixture will be x gram

Then,amount of NaHCO3 in the mixture will be (1-x) gram

molar mass of Na2CO3 = 106g/mol

therefore,

number of moles of Na2CO3 = x/106mol

molar mass of NaHCO3 = 84 g/mol

therefore,

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

ACCORDING TO THE QUESTION,

X/106 = 1-X/54

84x = 106 – 106x

190x = 106

x = 0.5579

therefore,

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

so,

1 mol of Na2CO3 reacts with 2 mol of HCl

therefore,0.0053 mol of Na2CO3 reacts with 2X0.0053 mol = 0.0106mol

and,

1 mol of NaHCO3 reacts with 1 mol of HCl

therefore,0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl

total moles of HCl = 0.0106 + 0.0053

= 0.0159 moles

In 0.1 mol HCL

0.1 mol of HCl is present in 1000ml of solution

therefore,

0.0159 mol of HCL is present in

1000 X 0.0159/0.1

= 159 ml of solution

Hence 159 ml of HCl is required to react completely with 1g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both

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