SOLUTION: You are mixing a 10% hydrochloric acid solution with 1% hydrochloric acid solution to make 12 L of a 4.5% solution. How much of each original solution should you use? HELP IM FAILI

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Question 569512: You are mixing a 10% hydrochloric acid solution with 1% hydrochloric acid solution to make 12 L of a 4.5% solution. How much of each original solution should you use? HELP IM FAILING AT THIS QUESTION.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = liters of 10% solution needed
Let +b+ = liters of 1% solution needed
+.1a+ = liters of h. acid in 10% solution
+.01b+ = liters of h. acid in 1% solution
given:
(1) +a+%2B+b+=+12+
(2) +%28+.1a+%2B+.01b+%29+%2F+12+=+.045+
----------------------------
(2) +.1a+%2B+.01b+=+.045%2A12+
(2) +.1a+%2B+.01b+=+.54+
(2) +10a+%2B+b+=+54+
Subtract (1) from (2)
(2) +10a+%2B+b+=+54+
(1) +-a+-+b+=+-12+
+9a+=+42+
+a+=+4.667+
and, since
(1) +a+%2B+b+=+12+
(1) +4.667+%2B+b+=+12+
(1) +b+=+12+-+4.667+
(1) +b+=+7.333+
4.667 liters of 10% solution are needed
7.333 liters of 1% solution are needed
check:
(2) +%28+.1a+%2B+.01b+%29+%2F+12+=+.045+
(2) +%28+.1%2A4.667+%2B+.01%2A7.333+%29+%2F+12+=+.045+
(2) +%28+.4667+%2B+.07333+%29+%2F+12+=+.045+
(2) +.54003+%2F+12+=+.045+
(2) +.54003+=+.54+
OK- error due to rounding off