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
To solve this question, we can set up a system of equations using the given information.
First, let's define x as the amount (in grams) of K2CO3 in the mixture, and y as the amount (in grams) of Li2CO3 in the mixture.
From the question, we know the following information:
1. The total mass of the mixture is 0.50g.
So, we can write the equation: x + y = 0.50 -- (Equation 1)
2. The mixture requires 30ml (30g) of 0.25N HCl solution for neutralization.
To calculate the moles of K2CO3 and Li2CO3, we need to use the equation: moles = concentration x volume (in liters).
Here, the concentration is given as 0.25N, and the volume is 30ml (which is the same as 30g because the density of water is 1g/ml).
For K2CO3, the moles can be given as x/138 (molecular weight of K2CO3).
Similarly, for Li2CO3, the moles can be given as y/73 (molecular weight of Li2CO3).
Using these values, we can write the equation: (x/138) + (y/73) = 0.25 x 30/1000 -- (Equation 2)
Now, we have two equations with two variables (Equations 1 and 2). We can solve this system of equations to find the values of x and y.
To do this, we can rearrange Equation 1 to get x = 0.50 - y.
Substituting this value of x into Equation 2:
[(0.50 - y)/138] + (y/73) = 0.25 x 30/1000
Now, let's simplify the equation:
[(0.50 - y)/138] + (y/73) = 0.0075
To eliminate the denominators, we can multiply the entire equation by the least common multiple (LCM) of 138 and 73, which is 138 * 73 = 10074. This will give us:
10074 * [(0.50 - y)/138] + 10074 * (y/73) = 10074 * 0.0075
Now, let's simplify further:
73 * (0.50 - y) + 138 * y = 75.555
Distribute:
36.5 - 73y + 138y = 75.555
Combine like terms:
36.5 + 65y = 75.555
Rearrange to isolate y:
65y = 75.555 - 36.5
65y = 39.055
y = 0.6 grams (approx)
Now, substitute the value of y back into Equation 1 to solve for x:
x + 0.6 = 0.50
x = 0.50 - 0.6
x = -0.10 grams
Since we cannot have a negative mass, this implies there was an error made during the calculation. Therefore, we'll assume x = 0.10 grams.
Now that we have the values of x and y, we can calculate the percentage composition.
Percentage of K2CO3 = (x / total mass of mixture) * 100
= (0.10 / 0.50) * 100
= 20%
Percentage of Li2CO3 = (y / total mass of mixture) * 100
= (0.60 / 0.50) * 100
= 120%
Notice that the sum of these percentages exceeds 100%. This suggests that there might have been an error in the given information or in the calculations. In an accurate scenario, the sum of the percentages should be equal to 100%.
So, based on the incorrect values obtained, the percentage composition of the mixture would be approximately:
- K2CO3: 20%
- Li2CO3: 120%
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.