Question

How many ml of 0.1 m hcl are required to react completely with 1g mixture of na2co3?

Answer 1

Answer:

Explanation:

HELLO RAVI,

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

THANK YOU

Answer 2

Let, amount of Na2CO3 in mixture be (x) g. Therefore, amount of NaHCO3 in mixture will be (1-x) g.

Molar Mass of Na2CO3 = 106 g/mol

Therefore, number of moles of Na2CO3 = x/106 mol

Molar Mass of NaHCO3 = 84 g/mol

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

Now A.T.Q.,

x/106 = (1-x)/84

ㅤㅤㅤㅤㅤㅤㅤ ㅤ84x = 106 - 106x

ㅤㅤㅤㅤㅤㅤㅤㅤ190x = 106

ㅤㅤㅤㅤㅤㅤㅤㅤ ㅤx = 0.5579

• So, No. of Moles of Na2CO3 = 0.0053 mol

ㅤㅤㅤㅤㅤㅤㅤalso, No. of Moles = 0.0053 mol

• REACTIONS :-

ㅤㅤㅤ2HCl + Na2CO3 → 2NaCl + H2O + CO2

ㅤ ㅤ(2 mol) ㅤㅤ(1 mol)

and, HCl + NaHCO3 → NaCl + H2O + CO2

ㅤㅤ(1 mol) ㅤ(1 mol)

• Since, 1 mol of Na2CO3 reacts with 2 mol of HCl.

ㅤㅤㅤ Therefore, 0.0053 mol of Na2CO3 reacts ㅤㅤㅤㅤwith (2 × 0.0053 = 0.0106) mol.

• Similarly, 0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl.

• TOTAL MOLES of HCl = 0.0106 + 0.0053 = 0.0159 mol.

• 0.1 mol of HCl is present in 1000 mL of solution.

ㅤㅤㅤTherefore, 0.0159 mol of HCl is present in ㅤㅤㅤㅤㅤSolution = 1000 × 0.0159/0.1

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ = 159 mL

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