There are two vessels containing 30 and 42 kg of solution. If they are drained together, 40% acid

There are two vessels containing 30 and 42 kg of solution. If they are drained together, 40% acid will be obtained, if equal masses are drained, then 37%. How much acid in 2 vessels?

1. The mass of the acid solution in the first vessel: M1 = 30 kg;

2. The percentage of acid in this solution: K1 %%;

3. The mass of the solution in the second vessel: M2 = 42 kg;

4. Percentage of acid: K2 %%;

5. In the third solution (when mixing the first and second) acids: K3 = 40%:

6. In the fourth solution (when mixing equal masses M = M1) acid: K4 = 37%;

7. Acid balance of the third solution:

K1 * M1 + K2 * M2 = K3 * (M1 + M2);

30 * K1 + 42 * K2 = 0.4 * (30 + 42) = 28.8;

8. Acid balance of the fourth solution:

K1 * M1 + K2 * M1 = K4 * (2 * M1);

30 * K1 + 30 * K2 = 0.37 * 2 * 30 = 22.2;

9. Subtract the second equation from the first:

K2 * (42 – 30) = 28.8 – 22.2 = 6.6;

K2 = 6.6: 12 = 0.55;

10. Acids in the second vessel:

M2k = K2 * M2 = 0.55 * 42 = 23.1 kg.

Answer: the second vessel contains 23.1 kg of acid.