Question

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?

Answer 1

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.

Answer 2

Answer:

158 mL

Explanation:

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:  

$Na_2CO_3 + 2Hcl \longrightarrow  2Nacl + H_2O + CO_2\\NaHCO_3 + Hcl \longrightarrow Nacl + H_2O + CO_2$

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 NaHCO₃, containing equimolar amounts of both.