SOLUTION: how much pure acid should be mixed with 7 gallons of a 40% acid solution in order to get a 70% acid solution?

Question 102385: how much pure acid should be mixed with 7 gallons of a 40% acid solution in order to get a 70% acid solution?
Found 2 solutions by stanbon, ptaylor:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how much pure acid should be mixed with 7 gallons of a 40% acid solution in order to get a 70% acid solution?
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Comment: Follow the acid.
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40% solution DATA:
Amt = 7 gallons ; amt of acid = 0.07*7 = 0.49 gallons
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100% solution DATA;
Amt = x gallos ; amt of acid = x gallons
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70% mixture DATA:
amt = x+7 gallons ; amt of acid = 0.70(x+7)= (0.70x + 4.9) gallons
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EQUATION:
acid + acid = acid
0.49 + x = 0.70x + 4.9
0.30x = 4.41
x = 14.7 gallons (amt of 100% solution that must be added)
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Cheers,
Stan H.

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of pure acid needed
Now we know that the amount of pure acid in the acid that's added (x) plus the amount of pure acid in the 40% solution ((7)*(0.40)) equals the amount of pure acid in the final mixture ((0.70)(7+x)). So our equation to solve is:
x+7*0.40=0.70(7+x) get rid of parens
x+2.8=4.9+0.70x subtract 0.70x and also 2.8 from both sides
x+2.8-2.8-0.70x=4.9-2.8+0.70x-0.70x collect like terms
0.30x= 2.1 divide both sides by 0.30
x=7 gallons---------------------------amount of pure acid needed
CK
7+7*0.40=0.70(14)
7+2.8=9.8
9.8=9.8
Hope this helps---ptaylor
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