# chemistry (Pls Bob check for me)

posted by
**Fai**
.

A 1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of s solution, 20cm3 of this solution required 40cm3 of 0.1N hcl for neutralization. Calculate the wieght of na2c03 and k2c03 in the mixture

The answer 0.4G

My own calculation:

na2c03 weight = 106g/2 = 53g/equiv

k2c03 weight = 138g/2 = 69g/equiv

x/53 na2c03 in 100ml

y/69 k2c03

0.04L x 0.1N = 0.004

x/53 + y/69 = 0.004

0.01886x + 0.014492(1.2g- x) = 0.004

0.01886x + 0.0173904 - 0.014492x = 0.004

0.01886x-0.01449x = 0.0173904 - 0.004

0.00437x = 0.01339

x =3 x 20/100

x = 0.612g

And then 1-0.612g

the answer = 0.4G

Bob check my answer.

Thank you for Bob