Posted by Fai on Friday, February 22, 2013 at 11:57pm.
0.50g of a mixture of K2c03 and Li2c03 requires 30ml of a 0.25N hcl solution for neutralization. What is the percentage composition of the mixture?
The answer k2c03 96%
Li2c03 4%
K2c03 weight 138g/mol
138/2 = 69
Li2c03 weight 74g/mol
74/2 = 37
30ml x 0.25N = 7.5 / 1000
x= 0.0075
Who helps me to solve it for me
Please Bob

chemistry  DrBob222, Saturday, February 23, 2013 at 1:07am
You have two unknowns; therefore, you must have two equations in the two unknowns.
Let X = mass K2CO3
and Y = mass Li2CO3

eqn 1:
X + Y = 0.500g
eqn 2:
(#Eq X) + (#Eq Y) = 0.0075 or
(X/69) + (Y/37) = 0.0075
Solve eqn 1 and 2 simultaneously for X an Y. Then
(X/0.5)*100 = %K2CO3
(Y/0.5)&100 = %Li2CO3