How many liters each of 15% acid and 33% acid should be mixed to get 120 L of 21% acid?

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Answer 1 · 2018-07-24T14:12:32

The volume of $15 %$ $15%$ is $= 80 L$ $=80L$ and the volume of $33 %$ $33%$ is $= 40 L$ $=40L$

Perform a material balance

The total volume is $= 120 L$ $=120L$

Let the volume of $15 %$ $15%$ acid be $= v L$ $=vL$

Then,

The volume of $33 %$ $33%$ acid is $= (120 - v) L$ $=(120-v)L$

The final percentage is $= 21 %$ $=21%$ acid

Therefore,

$v \cdot \frac{15}{100} + (120 - v) \cdot \frac{33}{100} = 120 \cdot \frac{21}{100}$ $v*15/100+(120-v)*33/100=120*21/100$

$15 v + 120 \cdot 33 - 33 v = 120 \cdot 21$ $15v+120*33-33v=120*21$

$18 v = 120 (33 - 21) = 1440$ $18v=120(33-21)=1440$

$v = \frac{1440}{18} = 80 L$ $v=1440/18=80L$