Question

When 15 g of a mixture of NaCl and Na,co, is heated with dilute HCI, 2.5 g of co, is evolved at NTP. Calculate percentage composition of the original mixture.


#follow me​

Answer 1

The reaction only involves sodium carbonate and hydrochloric acid and sodium chloride is only an impurity.

The equation is as follows: Na2CO3 + 2HCl--------> 2NaCl + H2O + CO2

2.5 grams of CO2 is evolved at NTP conditions.

molar mass of CO2= 44 grams per mole.

So, 2.5 grams of CO2 will be 2.5/44 moles.

=>2.5/44 moles of CO2 is evolved.

So, from the equation, 1 part of CO2 is evolved from one part of Na2CO3.

So, 1 part is equal to 2.5/44 moles => 0.0568 moles.

molar mass of Na2CO3 is 2*23+12+3*16 =>46+12+48 => 106 grams per mole.

So, 0.0568 moles of Na2CO3 has a mass of 106*0.0568 grams.

=> 6.0208 grams.

So, percentage composition of Na2CO3 in the 15 gram mixture:

6.0208/15*100 % => 40.14 %

Percentage composition of NaCl =100-40.14 => 59.86 %