Stochiometry help!!!!!!!!!?

Question

what mass of AgCl will precipitate when .050 L of a 0.05000 M solution of AgNO3 reacts with 25.0 mL of .0330 M Nacl

balanced equation:
AgNO3+NaCl=NaNO3+AgCl

please help me

Answer 1

Moles of AgNO3 = 0.050L * 0.05000 M = 2.5 x 10 -3 moles

moles NaCl = 0.025L * 0.0330 M = 0.825 x 10 -3 moles

So, this reaction would produce 0.825 x 10 -3 moles of AgCl

Mass = moles x molar mass (MM of AgCl = 143.32 g/mol)
= (0.825 x 10 -3 moles) x 143.32 g/mol
= 0.118 g
0.12 g AgCl will be precipitated

Answer 2

Hey - you need to know the Atomic weights of the different compounds. It's the other number besides the Atomic number on the period table (also called atomic mass). For each molecule, multiply the atomic weight of each atom, by the number of that particular atom - ex: AgNO3

1 x atomic mass of silver (Ag)
+
1 x atomic mass of Nitrogen (N)
+
3 x atomic mass if Oxygen (O - here, O3) === atomic mass of 1 molecule of AgNO3.

Now, get out your Avogadro Number (which I swear I can' t remember), but that is how many molecules of anything there are per mole of that thing.

Multiply as follows x moles substance per liter = xM solution

Then aM/L x avagadro #/mole of that substance - this cancels to leave you with a x avagadro/L = a molecules per liter - then just remember it's a matter of algebra - both sides of the equation must work out to the same number, so solve for the varialble. Easier to show on a chalkboard, really.

Answer 3

Well first let's find out how many moles of Silver Nitrate (AgNO3) we have:

.05L *.05moles/L=2.5x10^-3 moles AgNO3

And Salt (NaCl)
25mL=.025L
.025L*.033moles/L=8.25 x10^-4 moles NaCl

We must now find the limiting reactant. Since AgNO3 and NaCl react in a 1:1 ratio in terms of moles, the smaller number of moles of reactant will be the limiting reactant. Since

8.25 x10^-4 moles NaCl<2.5x10^-3AgNO3
then NaCl is the limiting reactant.

But notice for each mole NaCl, we get one mole AgCl. So for 8.25x10^-4 moles NaCl, we get the same number of moles AgCl.

So how much mass is 8.25x10^-4 moles of AgCl? First find the mass per mole of AgCl:

Ag(107.9g)+Cl(35.5g)=
143.4g/mole ofAgCl

We have
8.25x10^-4 moles * (143.4g/mole)= .12grams AgCl (assuming that AgCl is completely insoluble and precipitates fully)