How many liters of each should be mixed to give the acid needed for the experiment if it calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and 35%?

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Answer 1 · 2014-12-10T16:02:23

Let $X$ $X$ be the amount of 10% sulfuric acid that is required
Let $Y$ $Y$ be the amount of 35% sulfuric acid that is required

Since the resulting 15% sulfuric acid should be 1 L,

$Y = 1 - X$ $Y = 1 - X$

Equate the amount of pure sulfuric acid resulting from the parts and from the whole

$0.10 X + 0.35 (1 - X) = 0.15 (1)$ $0.10X + 0.35(1 - X) = 0.15(1)$

$\Rightarrow 100 [0.10 X + 0.35 (1 - X) = 0.15 (1)]$ $=> 100 [0.10X + 0.35(1 - X) = 0.15(1)]$

$\Rightarrow 10 X + 35 (1 - X) = 15 (1)$ $=> 10X + 35(1 - X) = 15(1)$

$\Rightarrow 10 X + 35 - 35 X = 15$ $=> 10X + 35 - 35X = 15$

$\Rightarrow - 25 X = 15 - 35$ $=> -25X = 15 - 35$

$\Rightarrow - 25 X = - 20$ $=> -25X = - 20$

$\Rightarrow X = \frac{20}{25}$ $=> X = 20/25$

$\Rightarrow X = 0.80$ $=> X = 0.80$

$Y = 1 - X$ $Y = 1 - X$

$\Rightarrow Y = 1 - 0.80$ $=> Y = 1 - 0.80$

$\Rightarrow Y = 0.20$ $=> Y = 0.20$

0.80 L of 10% Sulfuric Acid and 0.20 L of 35% Sulfuric Acid is required