Question

A hydrochloric acid solution consists of 28.0% HCl by mass and has a density of 1.14g/mL. This solution reacts completely with 1.36 g Al in the reaction 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g).
Determine the molarity of the AlCl3(aq) if the solution volume is simply the 17.3 mL.

Answer 1

Answer:

Explanation:

As you know, a

1.75-mol L

−

1

hydrochloric acid solution will contain

1.75

moles of hydrochloric acid, the solute, for every

1 L

of solution.

So right from the start, you know that your solution must contain

1.75

moles of hydrochloric acid.

Use the compound's molar mass to convert this to grams

1.75

moles HCl

⋅

36.46 g

1

mole HCl

=

63.805 g

Now, you know that your stock solution is

36.0

%

hydrochloric acid by mass, which implies that in order to have

36.0 g

of hydrochloric acid, you need

100 g

of this solution.

You can thus say that in order for your target solution to contain

63.805 g

of hydrochloric acid, the sample you take from the stock solution must have a mass of

63.805

g HCl

⋅

100 g solution

36.0

g HCl

=

177.24 g

Use the density of the stock solution to figure out the volume of the sample

177.24

g solution

⋅

1 mL

1.18

g solution

=

209 mL

−−−−−−−

The answer is rounded to three sig figs.

So what you would do is take

209 mL

of the stock solution and add it to enough water to get the total volume of the resulting solution to

1.00 L

.

Since you're working with concentrated hydrochloric acid, you should use an ice bath to cool the water before adding the stock solution to it.

You should also add the stock solution in multiple steps and stir the solution properly between each step.