[MaelStrom]

Alright Im typing up the lab report for a lab and would really appreaciate any *delete me*
In the lab we started with pure copper and cycled it into various forms through reactions. I am writing out all of the equations, but there is one equation which is giving me trouble.
I have aqueous copper chloride(CuCl2), and the aim is to regenerate pure copper. So we added pure magnesium(Mg). Since Mg is more active, a single replacement reaction occurs, and thus you get pure copper and MgCl2
Reaction: CuCl2(aq) + Mg(s) → Cu(s) + MgCl2(aq)
But in the lab, we added HCl afterwards to help along the process, to convert any unreacted Mg into Copper. My question is how does this work?Meaning, what would the equation be? I thought it might be something like this:
3Mg + Cu + CuCl2 + O2 + 4HCl =3MgCl2 + 2Cu + 2H20

I dont have a large amount of knowledge concerning reactions with acids, so I really need help with this one. Thanks in advance!

Edit: Not to sure why someone added *Delete me* into my post. I read the rules, and I dont believe I violated any of them. I made an attempt at solving the problem, or atleast thought about it as best as my education allows me, and wish for any reaffirmation of my attempt or a correction. Sorry if asking for help is not allowed in the "help" forum.

[MaelStrom]

Well I believe I have figured it out, thanks.
Single displacement
CuCl₂(aq) + Mg(s) → Cu²⁺(s) + MgCl₂(aq)
After adding HCl, reduction reaction
Mg(s) + Cu²⁺ (aq)+ CuCl₂ (aq)+ HCl(aq)→Cu(s) + MgCl₂(aq)+ H₂(g)

[MaelStrom]

Fin.
Single displacement reduction reaction
CuCl₂(aq) + Mg(s) → Cu(s) + MgCl₂(aq)
After adding HCl, single displacement reaction
Cu(s)+ MgCl₂(aq)+Mg(s) + 2HCl(aq)→ 2MgCl₂(aq)+ H₂(g)+Cu(s)
.

[Borek]

You are partially OK and partially wrong.

HCl was added not to facilitate copper reduction, but to get rid of the excess metallic magnesium. I guess your last step was filitration of preciptated metallic copper - you don't want metallic magnesium to be filtered and weigthed with copper.

Take a look at your reaction equations. If some substance is present on both sides, it cancels out - it simply didn't take part in the reaction. That will have especially strong effect on teh last reaction you wrote - cancel out substances that are present on both sides and the compare what is left with information from the previous paragraph.

[MaelStrom]

It took me a while to realize what the HCl was for, to clean up the mixture so to speak. Also, I wasnt sure if I was supposed to leave the specatator s in the equation, I thought it better that I acknowledge them, but I will remove them now. Thanks for the input!