# Question #26712

##### 1 Answer

Here's what I got.

#### Explanation:

The first thing to do here is to figure out exactly how much hydrochloric acid you get in

#25 color(red)(cancel(color(black)("mL solution"))) * "15 g HCl"/(100color(red)(cancel(color(black)("mL solution")))) = "3.75 g HCl"#

Now, if you take **number of grams** of hydrochloric acid coming from the **number of grams** of hydrochloric acid coming from the

#x + y = 3.75" "color(orange)((1))#

For the

#x color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(5color(red)(cancel(color(black)("g HCl")))) = (20x)" mL solution"#

For the

#y color(red)(cancel(color(black)("g HCl"))) * "100 mL solution"/(20color(red)(cancel(color(black)("g HCl")))) = (5y)" mL solution"#

You can now say that

#20x + 5y = 25" "color(orange)((2))#

Use equation

#x = 3.75 - y#

Plug this into equation

#20 * (3.75 - y) + 5y = 25#

#75 - 20y + 5y = 25#

#-15y = - 50 implies y = (-50)/(-15) = 10/3#

Use this to find the value of

#x = 3.75 - 10/3 = 15/4 - 10/3 = 5/12#

Therefore, you can say that in order to get

#20 * x = 20 * 5/12 = color(darkgreen)(ul(color(black)("8.3 mL of 5% HCl solution")))#

and

#5 * y = 5 * 10/3 = color(darkgreen)(ul(color(black)("16.7 mL of 20% HCl solution")))#

I'll leave the answers rounded to two and three **sig figs**, respectively, but keep in mind that you only have one significant figure for the concentrations of the two stock solutions.