Question

0.5 gram of a mixture of K2CO3 and Li2co3 required  30 ml of 0.25 M HCL solution for neutralization what is the percentage composition of mixture

Answer 1

Let X = mass K2CO3 
and Y = mass Li2CO3 
--------------------- 
eqn 1: 
X + Y = 0.500g 

eqn 2: 
(#Eq X) + (#Eq Y) = 0.0075 or 

equivalents K2CO3 + equivalents of Li2CO3 = 0.0075 or
(X/69) + (Y/37) = 0.0075 

0.01449X + 0.02703Y = 0.0075 
We can solve for Y from equation 1 by 

X + Y = 0.50 
Y = 0.50-X and substitute that into equation 2. 
(0.01449X) + 0.02703(0.50-X) = 0.0075 
0.01449X + 0.01352 - 0.02703X = 0.0075 
0.01449X - 0.02703X = 0.0075-0.01352 
-0.01254X = -0.00602 
X = 0.48g = mass K2CO3 
Then X + Y = 0.50 and 
0.48 + Y = 0.50 and 
Y = 0.02 mass Li2CO3 
%K2CO3 = (mass K2CO3/mass sample)*100 = 96% 
%Li2CO3 = (mass Li2CO3/mass sampl)*100 = 4%.