More help with concentrations?

Question

I can't get these type of problems. I need these in the most simplistic terms possible please. Thanks so much!!!!!

Calculate the volume of .23M sulfuric acid that would be needed to neutralize 146 mL of a 1.72 M aqueous ammonia solution. The equation for the reaction is H2SO4 + 2 NH3 --> 2 NH4 +SO4-2


Calculate the concentration of an aqueous KCl solution if 584 mL of the solution gives 16.9 g of AGCl when treated with excess AgNO3. The equation for the reaction is
KCl + AgNO3 --> AgCl + KNO3

A salt solution is evaporated to dryness. If 51.7 g of the solution leaves 1,725 mg of dry salt, what is the percent concentration (w/w) of the salt in the solution?
Thanks so much for any help!

Answer 1

1) 2 moles of NH3 need 1 mole of H2SO4

So, 146x1.72 mmol of MH3 needs (146x1.72)/2 mmoles of H2SO4

Vol. of .23M H2SO4 reqd. = (146x1.72)/(2x.23) ml

2) Molar mass of AgCl = 108+35.5 = 143.5 g

So, moles of AgCl = 16.9/143.5

1 mol of KCl gives 1 mol of AgCl

So 16.9/143.5 moles of KCl in 584 ml is equivalent to

16.9x1000/(143.5x584) M

3) 1.725 g is in 51.7 g of soln.

So, % conc is 1.725x100/51.7 = 172.5/51.7

I will leave it to you to do the math.

Answer 2

For the first question, you have a strong acid and a strong base. Thus, you need the same number of moles of H+ as NH3. For any stong acid/stong base question, you always need the same number of moles of acid as base. Since you have the concentrations of each solution, the question is pretty easy. You are starting with the volume of ammonia, and you want to end with the volume of sulfuric acid. The easiest way to do this is to simply multiply conversion factors.

(0.146L NH3)*(1.73mol/L)*(1 mol H2SO4/2 mol NH3)*(1 L/0.23 mol)= L of H2SO4 (don't have a calculator in front of me to give exact answer)

You would use a similar method in the second question. Since there is a one-to-one relationship between the KCl and AgCl, you know that you have the same number of moles of each. So, first you find the number of moles of AgCl by dividing the mass of your sample by the molecular weight. And then you would dividethat answer by your volume. That would give you the answer.

You will probably have to do lots of questions like these two if you are taking a chemistry course. The trick is to realize that you do almost exactly the same thing for every question (as long as the reaction goes to completion). Once you understand how to do one problem, you will be able to do any problem like it.