Tuesday
September 23, 2014

Homework Help: chemistry (Pls Bob check for me)

Posted by Fai on Sunday, February 24, 2013 at 4:09pm.

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

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