A lab assistant wants to make 5 liters of 27.2% acid solution. If solutions of 30% and 16% are in stock, how many liters of each must be mixed to prepare the solution?
amount of the 30% solution needed ---- x L
amount of the 16% solution needed --- 5-x L
.3x + .16(5-x) = .272(5)
times 100
30x + 16(5-x) = 136
30x +80 - 16x = 136
14x = 56
x = 4 L
4 L of the 30%, and 1 L of the 16% solution would be needed.
To find the number of liters of each solution needed, we can set up a system of equations based on the given information.
Let's assume:
- x liters of the 30% acid solution
- y liters of the 16% acid solution
Since we want to make a 5-liter solution, we know that:
x + y = 5 ----(equation 1)
We also know that the 27.2% acid solution will be a combination of the 30% and 16% acid solutions. The amount of acid in 30% solution is 0.3x liters and the amount of acid in 16% solution is 0.16y liters.
To calculate the amount of acid in the final solution, we multiply the percentages by the total volume (5 liters) and set it equal to 27.2% of the total volume:
0.3x + 0.16y = 0.272 * 5
0.3x + 0.16y = 1.36 ----(equation 2)
Now we have a system of equations:
x + y = 5 ----(equation 1)
0.3x + 0.16y = 1.36 ----(equation 2)
To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method.
From equation 1, we can express x in terms of y as:
x = 5 - y
Substituting this value into equation 2:
0.3(5-y) + 0.16y = 1.36
Simplifying:
1.5 - 0.3y + 0.16y = 1.36
0.14y = 1.36 - 1.5
0.14y = -0.14
y = -0.14 / 0.14
y = -1
Now, since y represents the number of liters of the 16% acid solution, we can see that it doesn't make sense to have a negative amount. This tells us that the initial assumption is incorrect.
Let's try the elimination method instead.
Multiply equation 1 by 0.16 and equation 2 by 100 to eliminate decimals:
0.16x + 0.16y = 0.16 * 5
30x + 16y = 136
Now, we have:
0.16x + 0.16y = 0.8 ----(equation 3)
30x + 16y = 136 ----(equation 4)
Subtract equation 3 from equation 4:
(30x + 16y) - (0.16x + 0.16y) = 136 - 0.8
29.84x = 135.2
x = 135.2 / 29.84
x ≈ 4.53
Now, we have found that x ≈ 4.53 liters of the 30% acid solution and y ≈ 5 - x ≈ 5 - 4.53 ≈ 0.47 liters of the 16% acid solution.
So, the lab assistant needs approximately 4.53 liters of the 30% acid solution and 0.47 liters of the 16% acid solution to prepare the desired 5 liter 27.2% acid solution.