How do I solve this, please be detailed

Question

 

Zinc chloride (0.55 M, 50.0 mL) reacts completely with the ammonium phosphate(0.75 M, 55.0 mL). 

1. Write a balanced chemical equation describing the reaction
2. What mass of zinc phosphate is produced 
3. Assuming the volumes are additive, what is the final concentration of ammonium chloride? 
4. What is the access we can’t, and what is its final concentration? 

Answer 1

a) 3ZnCl₂ + 2(NH₄)₃(PO₄) ---->. 6(NH₄)Cl + Zn₃(PO₄)₂

To balance equation, make sure you keep in mind correct charges of anions/cations to get right molecular formulas.

b) First we need to find limiting reagent. Will use equation Molarity*volume = moles to convert into mmol.

0.55 M ZnCl₂ * 0.05 L ZnCl₂ = 0.0275 mol ZnCl₂

0.75 M (NH₄)₃(PO₄) * 0.055 L (NH₄)₃(PO₄) = 0.04125 mol (NH₄)₃(PO₄)

Using the mol ratio from the balanced equation in a, we can see how much ammonium phosphate is needed to react all of the zinc chloride.

0.0275 mol ZnCl₂ • 2 mol (NH₄)₃(PO₄)/3 mol ZnCl₂ = 0.0183 mol (NH₄)₃(PO₄)

Since we have more than enough ammonium phosphate, the zinc chloride is limiting. Knowing this we get the following equation:

0.0275 mol ZnCl₂ • 1 mol (Zn)₃(PO₄)₂/3 mol ZnCl₂ • 386.11 g (Zn)₃(PO₄)₂/1 mol (Zn)₃(PO₄)₂ = 3.54 g zinc phosphate.

c) first we have to find how many mol of ammonium chloride we make. We still know Zinc chloride is limiting, so we get

0.0275 mol ZnCl₂ • 6 mol (NH₄)Cl/3 mol ZnCl₂ = 0.055 mol ammonium chloride

since we can assume volumes are additive, the total volume is 105 mL or 0.105 L

0.055/0.105 = 0.524 M ammonium chloride.

d) there's a typo in the question, but I'm assuming the question is "what is the excess reagent and what is it's final concentration?"

in that case, we already know that ammonium phosphate is excess from part b. We also know that we use 0.0183 mol of it for the reaction from part b.

0.04125 mol (NH₄)₃(PO₄) initial - 0.0183 mol (NH₄)₃(PO₄) consumed = 0.02295 mol (NH₄)₃(PO₄) remaining

The total volume is still 0.105 L so the concentration is 0.02295/0.105 = 0.219 M ammonium phosphate