If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L HCL(aq) and 0.124 mol/L AgNO3(aq) what volume in litres of AgNO3 did we
add?

Approximately (rounded to three significant figures) assuming that is in excess.

Explanation:

When and precipitate, (the said precipitate) and are produced:

(verify that this equation is indeed balanced.)

Look up the relative atomic mass of and on a modern periodic table:

Calculate the formula mass of the precipitate, :

.

Calculate the number of moles of formula units in of this compound:

.

Notice that in the balanced equation for this reaction, the coefficients of and are both one.

In other words, if (the other reactant) is in excess, it would take exactly of formula units to produce of formula units.

Hence, it would take of formula units to produce of formula units.

Calculate the volume of the solution given that the concentration of the solution is :

.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of .)

In other words, approximately of that solution would be required.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.