How many liters of a 80% acid solution must be added to 12 liters of a 10% acid solution to produce a 38% acid solution?

Answer 1

8 Liters

80E + 10(12) = 38(E+12)

80E +120 = 38E + 456

42E = 336

7E = 56

E = 8 liters

80(8)+ 10(12) = (8+12)(38)

640 + 120 = (20)(38)

760 = 760