Can someone please explain how to answer stoichiometry problems?

Question

I have this worksheet that's supposed to help, but it doesn't. One of the questions says: How many grams of aqueous iron(II) bromide will be needed to react with an excess of potassium phosphate, K3PO4, to produce 17.1 grams of solid iron(II) phosphate? I know you have to set up the equation and balance it. I don't know if this is right, but i wrote: FeBr2+K3PO4>Fe(PO4)2 so could someone please try to help me with the rest of the problem and help with my formula because I don't think it's right :(

Answer 1

First you need to correctly balance the chemical equation:

3FeBr2 + 2K3(PO4)2 ------ Fe3(PO4)2 + 6KBr

This is a double replacement reaction in which the positive ions trade places to produce two new compounds. The charge of each ion is as follows:

Fe +2
Br -1
K +1
PO4 -3
Chemical formulas are balanced and then the chemical equation.

From there you are doing a mass-mass stoichiometry problem.
1. Take the mass of the iron (II) phosphate given divided by the molar mass of iron(II) phosphate from the periodic table.
2. Multiply by the mole ratio of iron (II) bromide / iron (II) phosphate from your balanced equation
3. Multiply by the molar mass of iron (II) bromide.

1. 17.1 g Fe3(PO4)2 / 357.49 g/mol Fe3(PO4)2 = .048 moles
2. .048 moles Fe3(PO4)2 X 3FeBr2 / Fe3(PO4)2 = . 144 moles FeBr2
3. .144 moles FeBr2 X 215.65 g/mole FeBr2 = 31.1 g FeBr2

Answer 2

3FeBr2 + 2K3PO4 ===> Fe3(PO4)2 + 6KBr

You need to know the concentration of FeBr2 solution. Also, I think the question asks for the mL of the solution, not grams.

Answer 3

For this type of problem, the main goal is to find which reactant is the "limiting reactant", or which one will run out first. You know you need 3 mol of Si and 2 mol of N2 to produce 1 mol of product. So the first thing you do is figure out how many moles of the product you could make with both of your samples. First for N2 0.62 / 2 = 0.31 (moles divided by number of moles needed for 1 mol product) So if you had unlimited Si, you could make 0.31 mol Si3N4 with your given amount of N2. Then for Si 0.75 / 3 = 0.25 Since you can produce a smaller amount of product with the Si, it is the limiting reactant, after you make 0.25 mol of your product, you will have no Si left and have an excess of N2.