Help finding grams from reaction?

Question

What volume (in mL) of 0.127 M AgNO3 is required to react completely with 33.92 mL of 0.0750 M FeBr2?

2 AgNO3 + FeBr2 → 2 AgBr + Fe(NO3)2

I got 40.0 mL as the volume. Is this right?

2)How many grams of AgBr are formed?

Answer 1

Moles(FeBr2) = 0.075 x 33.92/1000 = 0.002544
From the molar ratios 2 moles of AgNO3 are required to react with 1 mole of FeBr2.
So Mol(AgNO3) = 0.002544 x 2= 0.005088
Using and rearranging the equation:-

moles = [conc] x volume / 1000cm^3
volume/cm^3 = moles x 1000 cm^3/ [conc]

So
volume = 0.005088 x 1000 / 0.127 = 40.063 cm^3 ~ 40 cm^3

Yes!!! your answer is correct.

2)
Mass (AgBr)
As molar ratios between AgNO3 & AgBr are 2:2
then 0.005088 moles AgNO3 produces 0.005088 moles(AgBr).
Mr(AgBr) =
Ag x 1 = 108 x 1 = 108
Br x 1 = 80 x 1 = 80
108 + 80 = 188 (Mr(AgBr))
mass(AgBr) = 0.005088 x 188 = 0.956544 g

Answer 2

1) moles FeBr2 = CV = (0.0750 M)(0.03392 mL) = 0.00254 moles

According to balanced reaction 2 moles of AgNO3 reacts with 1 mol of FeBr2:

2 moles AgNO3 ---- 1 mol FeBr2
.......x ---------- 0.00254 mol FeBr2
x = 0.00508 moles AgNO3

Volume of AgNO3 that contains this amount of reactant:

0.127 moles AgNO3 -- 1 L
0.00508 moles ------ x
x = 0.04 L = 40 mL

You are right!

Well done!

Regarding the second question:

1 mol FeBr2 --- 2 moles AgBr
0.00254 mol FeBr2 ---x
x = 0.00508 moles AgBr2

Molecular weight of AgBr2 = 187.77 g/mol hence:
mass of Br = (0.00508 moles)(187.77 g/mol) = 0.9538 g

Good luck!

Answer 3

33.92mLFeBr2 x 0.0750molFeBr2/1000mLFeBr2 x 2molAgNO3/1molFeBr2 x 1000mLAgNO3/0.127molAgNO3 = 40.06mL AgNO3

Atomic weights: Ag=108 Br=80 AgBr=188

33.92mLFeBr2 x 0.0750mLFeBr2/1000mLFeBr2 x 2molAgBr/1molFeBr2 x 188gAgBr/1molAgBr = 0.9565g AgBr