Pure acid is too be added to a 10% acid solution to obtain 54 L of a 20% acid solution.
fill this in
___L of pure acid solution and ___L of 10% acid solution.
What are the missing amounts from that?
+23254 A formula that works for problems like this is:
(Percent)·(Amount) + (Percent)·(Amount) = (Final Percent)·(Final Amount)
Since the final amount is 54L, the amounts that are combined to make this are: x and 54 - x.
Let the amount of pure acid to be x and the amount of the 10% acid solution to be 54 - x.
Pure acid is 100% acid.
(Pure acid) + (10% Solution) = (20% solution)
(1.00)(x) + (0.10)(54 - x) = (0.20)(54)
x + 5.4 - 0.1x = 10.8
0.9x + 5.4 = 10.8
0.9x = 5.4
---> x = 6 L of pure acid to be added to 48 L of a 10% solution.
+23254 A formula that works for problems like this is:
(Percent)·(Amount) + (Percent)·(Amount) = (Final Percent)·(Final Amount)
Since the final amount is 54L, the amounts that are combined to make this are: x and 54 - x.
Let the amount of pure acid to be x and the amount of the 10% acid solution to be 54 - x.
Pure acid is 100% acid.
(Pure acid) + (10% Solution) = (20% solution)
(1.00)(x) + (0.10)(54 - x) = (0.20)(54)
x + 5.4 - 0.1x = 10.8
0.9x + 5.4 = 10.8
0.9x = 5.4
---> x = 6 L of pure acid to be added to 48 L of a 10% solution.
