A chemist has two solutions of sulfuric acid. The first is one half sulfuric acid and half water. The second is three-fourths sulfuric acid and one fourths water. He wishes to make 10 liters of a solution which is two-thirds sulfuric acid and one-thirds water. How many liters of each available solution should he use?
Please be as detailed as you can i really want to understand this problem more than getting the solution THANKS^^
add up the amounts of acid in each portion:
1/2 x + 3/4 (10-x) = 2/3 * 10
x = 10/3
so, 10/3 L of 50% acid
and 20/3 L of 75% acid
To solve this problem, we can set up a system of equations.
Let's denote:
x = liters of the first solution (half sulfuric acid and half water) to be used
y = liters of the second solution (three-fourths sulfuric acid and one-fourth water) to be used
We are given that the chemist wants to make 10 liters of a solution that is two-thirds sulfuric acid and one-third water. From this information, we can establish the following equation:
x + y = 10 ... (Equation 1)
Additionally, we know that the first solution is one-half sulfuric acid and half water, while the second solution is three-fourths sulfuric acid and one-fourth water. To calculate the amount of sulfuric acid and water in the final solution, we can use the following equations:
Sulfuric acid in the first solution: (1/2)x
Water in the first solution: (1/2)x
Sulfuric acid in the second solution: (3/4)y
Water in the second solution: (1/4)y
In the final solution, we want a ratio of two-thirds sulfuric acid and one-third water. Therefore, we can set up the following equations:
Sulfuric acid in the final solution: (2/3)(x + y)
Water in the final solution: (1/3)(x + y)
Now, we can set up another equation based on the sulfuric acid and water content:
Sulfuric acid equation: (1/2)x + (3/4)y = (2/3)(x + y) ... (Equation 2)
To solve this system of equations (Equation 1 and Equation 2), we can use substitution or elimination method. Let's use the elimination method:
Multiply Equation 1 by 2/3 to make the equation compatible with the second equation:
(2/3)(x + y) = (2/3)(10)
(2/3)x + (2/3)y = 20/3
Now, Equation 2 becomes:
(1/2)x + (3/4)y = (2/3)x + (2/3)y
Multiply the equation by 12 to eliminate fractions:
6x + 9y = 8x + 8y ... (Equation 3)
Now, we can combine like terms in Equation 3:
6x - 8x + 9y - 8y = 0
-2x + y = 0
Now, we can solve this equation for either x or y. Let's solve it for x:
-2x = -y
x = (1/2)y ... (Equation 4)
Since we have the value of x in terms of y, we can substitute this value into Equation 1:
(1/2)y + y = 10
(3/2)y = 10
y = (10 * 2) / 3
y = 20/3
y = 6.67 liters (approximately)
Now, we can substitute the value of y back into Equation 4 to find x:
x = (1/2)(6.67)
x = 3.33 liters (approximately)
Therefore, the chemist should use approximately 3.33 liters of the first solution (half sulfuric acid and half water) and approximately 6.67 liters of the second solution (three-fourths sulfuric acid and one-fourth water) to obtain a final solution of 10 liters, which is two-thirds sulfuric acid and one-third water.