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- Thread starter jenn2998
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Picture 3 vats. One vat has x liters of 18% acid. The second vat has y liters of 45% acid. The third vat is empty, but will contain 12 liters of the new mixture.
Build two equations. One equation is based on how many liters of the 18% solution is combined with the number of liters of the 45% solution to get 12 liters of the new mixture.
The second equation is simply the amount of acid in the 18% solution plus the amount of acid in the 45% solution equals the amount of acit in the 36% solution. Hint: The amount of acid in the 18% solution is 0.18x.
If you need more help, show us how far you got, so we know where you are stuck.
Build two equations. One equation is based on how many liters of the 18% solution is combined with the number of liters of the 45% solution to get 12 liters of the new mixture.
The second equation is simply the amount of acid in the 18% solution plus the amount of acid in the 45% solution equals the amount of acit in the 36% solution. Hint: The amount of acid in the 18% solution is 0.18x.
If you need more help, show us how far you got, so we know where you are stuck.
Hello, jenn2998!
Here's how I baby-talk my way through these Mixture Problems . . .
Consider the amount of acid in the solutions.
\(\displaystyle \text{Let } x\text{ = amount of the 18 percent solution.}\)
. . \(\displaystyle \text{It contains: }\:\text{18 percent} \times x \:=\:0.18x \text{ liters of acid.}\)
\(\displaystyle \text{Then }12-x\text{ = amount of the 45 percent solution.}\)
. . \(\displaystyle \text{It contains: }\:\text{45 percent} \times (12-x) \:=\:0.45(12-x)\text{ liters of acid.}\)
. . \(\displaystyle \text{The final mixture contains: }\:0.18x + 0.45(12-x)\text{ liters of acid.}\) .[1]
\(\displaystyle \text{But we know that the final mixture will be: 12 liters which is 36 percent acid.}\)
. . \(\displaystyle \text{It will contains: }\:0.36 \times 12 \:=\:4.32\text{ liters of acid.}\) .[2]
\(\displaystyle \text{We just described the final amount of acid in }two\text{ ways.}\)
\(\displaystyle There\text{ is our equation! }\quad\hdots\quad 0.18x + 0.45(12-x) \:=\:4.32\)
Go for it!
Here's how I baby-talk my way through these Mixture Problems . . .
A chemist needs to mix a 18% acid solution with a 45% acid solution
to obtain 12 liters of a 36% acid solution.
How many liters of each of the acid solutions must be used?
Consider the amount of acid in the solutions.
\(\displaystyle \text{Let } x\text{ = amount of the 18 percent solution.}\)
. . \(\displaystyle \text{It contains: }\:\text{18 percent} \times x \:=\:0.18x \text{ liters of acid.}\)
\(\displaystyle \text{Then }12-x\text{ = amount of the 45 percent solution.}\)
. . \(\displaystyle \text{It contains: }\:\text{45 percent} \times (12-x) \:=\:0.45(12-x)\text{ liters of acid.}\)
. . \(\displaystyle \text{The final mixture contains: }\:0.18x + 0.45(12-x)\text{ liters of acid.}\) .[1]
\(\displaystyle \text{But we know that the final mixture will be: 12 liters which is 36 percent acid.}\)
. . \(\displaystyle \text{It will contains: }\:0.36 \times 12 \:=\:4.32\text{ liters of acid.}\) .[2]
\(\displaystyle \text{We just described the final amount of acid in }two\text{ ways.}\)
\(\displaystyle There\text{ is our equation! }\quad\hdots\quad 0.18x + 0.45(12-x) \:=\:4.32\)
Go for it!