Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

How many mL of 0.1M HCl is required to react completely with 1g of mixture of Na2CO3 and NaHCO3 containing equi-molar amounts of both ?

How many mL of 0.1M HCl is required to react completely with 1g of mixture of Na2CO3 and NaHCO3 containing equi-molar amounts of both ?

Grade:11

2 Answers

varun
304 Points
4 years ago
HELLO RAVI,
Let the amount of Na2CO3 in the mixture will be 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
 
Ajeet Tiwari
askIITians Faculty 86 Points
one year ago
hello student

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:



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.

hope it helps
thankyou


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free