1. Stoichiometry(Limiter, Percent Yield) Help

Hi everyone, I'm having trouble with the following questions
Reaction: SrCl2+CuSO4 SrSO4 + CuCl2

25.0 mL of .1 molar strontium chloride and 0.99 g of copper.

Here are the Questions:
1) Determine which reactant is limiting and what mass of the other reactant should remain after the reaction is complete.
2) Determine the theoretical yield that was expected.
3) Write a net ionic equation for this reaction.

My Attempt at the question:

Moles of SrCl2 = 0.0025 (0.025X0.1)

Moles of Cu(SO4)
= m/mm
= 0.99/159.56
=0.0062 mols

This would make SrCl2 the limiter in the reaction. Is that right?

SrSO4 is the solid that forms...
1 mol CuSO4 reacts to make 1 mol SrSO4
So the mols of excess reagent left over would be 0.0062-0.0025 = 0.0037

m = n x mm
= (0.0037 x 159.6)
= 0.59g of the excess left over.

Does that sound right?

2) 1 mol of SrCl2 reacts to make 1 mol SrSO4
0.0025 mol x
x= 0.0025 mol
m = n x mm
= (0.0025x183.7)
= 0.46 g of SrSO4

3) No idea how to make the net ionic equation

Sorry if my answers are a little unclear I'm not exactly sure if I'm doing the right thing. Thanks everyone

Also, If you need me to clarify anything just lemme know

2. They all seem good to me. (I haven't checked the relative formula masses though)

The ionic equation will be:

$\displaystyle SrCl_2 (aq) + CuSO_4 (aq) \rightarrow SrSO_4 (s) + CuCl_2 (aq)$

Write the state symbols. These will help you in such situations. Aqueous compounds will be in the form of ions, and solid compounds will remain as such.

$\displaystyle Sr^{2+} (aq)+ 2Cl^- (aq)+ Cu^{2+} (aq)+ SO_4\ ^{2-}(aq) \rightarrow SrSO_4 (s)+ Cu^{2+}(aq) + 2Cl^-(aq)$

Cross out the ions appearing on both sides:

$\displaystyle Sr^{2+} (aq)+SO_4\ ^{2-}(aq) \rightarrow SrSO_4(s)$

There you are!