Chemistry, asked by priyadharshini9683, 1 year ago

When 15g of a mixture of nacl and na2co3 is heated with dilute hcl 2.5g of co2 is evolved at ntp.calculate the percentage composition of the original mixture?

Answers

Answered by danielochich
14
Only Na2CO3 will react in this case


Na2CO3 + 2 HCl = 2 NaCl + CO2 + H2O

Molar mass of Na2CO3 = 106

Molar mass of CO2 = 44

44g of CO2 is produced from 106g Na2CO3

What about 2.5g of CO2 = (2.5 x 106)/44

                                        = 6.023 g


The original mixture has 6.023 g of Na2CO3

Percentage of Na2CO3 = 6.023/15 x 100 = 40.15%


Percentage of NaCl = 100  - 40.15 = 59.85%
Answered by Anonymous
3

Answer:

Explanation:

Only Na2CO3 will react in this case

Na2CO3 + 2 HCl = 2 NaCl + CO2 + H2O

Molar mass of Na2CO3 = 106

Molar mass of CO2 = 44

44g of CO2 is produced from 106g Na2CO3

What about 2.5g of CO2 = (2.5 x 106)/44

= 6.023 g

The original mixture has 6.023 g of Na2CO3

Percentage of Na2CO3 = 6.023/15 x 100 = 40.15%

Percentage of NaCl = 100 - 40.15 = 59.85%

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