SOLUTION: How many liters of a 35% acid solution should be mixed with a 14% solution to obtain 70 liters of a 20% acid solution? A chemist wants to obtain 15 liters of 24% alcohol solutio

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Question 95599: How many liters of a 35% acid solution should be mixed with a 14% solution to obtain 70 liters of a 20% acid solution?
A chemist wants to obtain 15 liters of 24% alcohol solution by combining a quantity of 20% solution, a quantity of 30% alcohol solution and 1 liter of pure water. How many liters of each of the alcohol solutions must be used?
How many kg each of scrap iron and steel containing 75% iron must be used to produce 15 kg of 8% steel rod?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
(1) How many liters of a 35% acid solution should be mixed with a 14% solution to obtain 70 liters of a 20% acid solution?
Let x=amount of 35% solution needed
Then70-x=amount of 14% solution needed
Now we know that the amount of pure acid in the 35% solution (0.35x) plus the amount of pure acid in the 14% solution (0.14(70-x)) has to equal the amount of pure acid in final mixture (0.20(70)). So our equation to solve is:
0.35x+0.14(70-x)=0.20(70) get rid of parens
0.35x+9.8-0.14x=14 subtract 9.8 from both sides
0.35x-0.14x+9.8-9.8=14-9.8 collect like terms
0.21x=4.2 divide both sides by o.21
x=20 liters--------------------amount of 35% solution needed
70-x=70-20=50 liters --------------amount of 14% solution needed
CK
0.35*20+0.14*50=0.20*70
7+7=14
14=14
(2) A chemist wants to obtain 15 liters of 24% alcohol solution by combining a quantity of 20% solution, a quantity of 30% alcohol solution and 1 liter of pure water. How many liters of each of the alcohol solutions must be used?
Amount of alcohol solutions plus the water =15 liters
Let x=the amount of 20% solution needed
Then 14-x=amount of 30% solution needed
Now we know that the amount of pure alcohol in the 20% solution (0.20x) plus the amount of pure alcohol in the 30% solution (0.30(14-x) plus the amount of pure alcohol in the water (0) equals (0.24*15). So our equation to solve is:
0.20x+0.30(14-x)=0.24*15 get rid of parens
0.20x+4.2-0.30x=3.6 subtract 4.2 from both sides
0.20x+4.2-4.2-0.30x=3.6-4.2 collect like terms
-0.10x=-0.6 divide both sides by -0.10
x=6 liters ------------------------amount of 20% solution needed
14-x=14-6=8 liters ---------------------amount of 30% solution needed
CK
0.20*6+0.30*8=0.24*15
1.2+2.4=3.6
3.6=3.6

(3) How many kg each of scrap iron and steel containing 75% iron must be used to produce 15 kg of 8% steel rod?

Let x=amount of steel containing 75% iron (and 25% steel) that's needed
Then 15-x=amount of pure iron (containing 0% steel) that's needed
Now we know that the amount of pure steel (0.25x)in the 75% alloy plus the amount of pure steel in the scrap iron (0) has to equal the amount of pure steel in the final steel rod (0.08(15)). So our equation to solve is:
0.25x=0.08*15 or
0.25x=1.2 divide both sides by 0.25
x=4.8 kg-------------------------amount of steel containing 75% iron that's needed
15-x=15-4.8=10.2 kg--------amount of scrap iron needed
Ck
0.25*4.8+0*10.2=0.08*15
1.2=1.2

Hope this helps----ptaylor