Chem help!!??!!?

Question

Statement: When 2.0 ml of 1.50-M Ammonium Phosphate is mixed with 1.000 ml of .750-M Ferrous Acetate, a precipitate forms.

Question: Calculate the theoretical yield (in grams) of the precipitate and identify the limiting reactant.

Please help!! I've been stuck on this for so long.

Answer 1

First you need the balanced equation for the reaction:

2(NH4)3PO4 + 3Fe(C2H3O2)2 --> Fe3(PO4)2 + 6(NH4)(C2H3O2)

from this we see that 2 moles of the Ammonium compound react with 3 moles of the Iron compound.

2.0 mL = 0.002 L
0.002 L * 1.5 mole/L = 0.003 moles

1.00 mL = 0.001 L
0.001 L * 0.750 mole/L = 0.00075 moles

The limiting reactant is the Iron (II) Acetate.

Divide the number of moles of Iron (II) Acetate by three to get the mole of product. Multiple this by the formula (molecular) weight of the product to give grams of product.

Answer 2

Ferrous acetate = iron (II) acetate
Fe (CH3COO)2

the precipitate is ferrousphosphate Fe3(PO4)2
obviously Ferrous acetate is the limiting reactant since it is both lower in amount and concentration AND the equation requires 3 mols of it to 2 mols of phosphate

proceed as thus

Answer 3

First, write out a balanced equation, making sure to use the correct chemical formulas for each substance.

The rest is stoichiometry.