Question 585459: A chemistry teacher needs to make 10L of 42% sulphuric acid solution. The acid solutions available are 30% sulphuric acid and 50% sulphuric acid, by volume. How many liters of each solution must be mixed to make the 42% solution?
My question is how do I set up the two system equations so I can then use the substitution method to solve.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! STEP 1:
Set variable names for the volumes you want to find.
Let x be the volume of 30% acid solution used, in L.
Let y be the volume of 50% acid solution used, in L.
(The volume of final mixture is given in L, and we want the units for all volumes to match).
STEP 2:
Write an equation that accounts for the total volume as a function of your variables. Assume volumes are addititve, meaning that when you mix a volume x of one solution with a volume y of amother solution the volume of the mix is x+y. (Chemists know that this is not exactly true, although usually the difference does not matter for your purpose).
 with all volumes in L
STEP 3:
Write an equation that accounts for the total amount of acid as a function of your variables. Be aware that 42% of some amount is 0.42 times that amount. Smilarly, 30% of some amount is 0.30 times that amount, and 50% of some amount is 0.50 times that amount.
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NOTES (optional reading):
In math 0.30 = 0.3 and 0.50 = 0.5, and you see the zeros as unneeded, but in science those zeros mean something about the precision of the measurements. If this is science homework, keep those zeros. If it's math, feel free to leave them out.
As long as your units for volume and concentration match, do not worry about what the units for amount are. They could be L if the percentages were volume in volume (% v/v), or they could be kilograms, if the percentages were weight in volume (w/v). They are the same, whatever they are.
THE SYSTEM
With those two equations, your system is:
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