You don't have one equation; you have two. You need two equations when you have two unknowns. Solve them simultaneously.
(X = mass K2CO3 and Y = mass Li2CO3)
eqn 1 is X + Y = .500 from the problem.
eqn 2 is (X/69) + (Y/37) = 0.0075
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explanations:
eqn 1. The problem tells you that the mixture of K2CO3 and Li2CO3 = 0.500; therefore, X(K2CO3) + Y(Li2CO3) = 0.50g
eqn 2. How many equivalents do you have? You have L x N = 0.03L x 0.25N = 0.0075. Therefore, equivalents K2CO3 + equivalents of Li2CO3 = 0.0075.
How many equivalents of K2CO3 do you have. That's X grams K2CO3/equivalnt weight = X/69.
How many equivalents of Li2CO3 do you have? That's grams Li2CO3/equivalent wight = Y/37. Total equivalents = 0.0075
That's where equation comes from.
(X/69) + (Y/37) = 0.0075.
Solve those two equations simultaneously. I will start but this is not a math class you are taking. You should know how to do this.
X + Y = 0.500
(X/69) + (Y/37) = 0.0075
We can simplify equn 2 this way.
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%.
Most of the questions you've asked recently I've had to work step by step including the algebra.
It appears to me that it's easier to say you don't understand than it is to work through the algebra. I have told you, step by step, how to solve the chemistry part. The rest is simple algebra.
0.50g of 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%
The answer li2c03 4%
I want Bob step by step explain. I do not still understand
x/69 + y/37 = 0.0075 How to calculate
1 answer
1 answer