Question

Mg+AlCL3---->Al+MgCl2 my bad

Answer 1

Ok, so what's the basic principle here again?

The number of atoms of each type of element must be equal on both sides of the reaction:
- on the left hand side, LHS, the reactants
- on the right hand side, RHS, the products

Answer 2

yes

Answer 3

So let's try this again. Let's make a guess of 1 Mg atom of reactants, the LHS

1 Mg+ ______ AlCL3 ----> ____ Al + _____ MgCl2

Now, how many molecules of MgCl2 must that result in?

Answer 4

Ok just first get it done in the way james is suggesting, and once done I will post the algebraic method

Answer 5

2MgCl right

Answer 6

or would it be 1MgCl because there is Cl3 on the LHS and then there is Cl2 on the RHS so you need one more Cl to balance it and then you wouldnt need any more Mg so just MgCl

Answer 7

Yes, one MgCl2 is going to cause problems because we need to balance the Cl3 in the reactants.

But this works:

3 Mg+ ______ AlCL3 ----> ____ Al + 3 MgCl2

Now we have 6 Cl atoms on the RHS, the products. Can you fill the rest now?

Answer 8

i think so you would have to put a 2 in front of the AlCl3 on the LHS, but as an effect you would have 2Al on the LHS so then i would put a 2 in front of the Al on the RHS and it would make 3Mg+2AlCl3--->2Al+3MgCl2

Answer 9

yes, that's it.

Answer 10

Now 2b will give you a less grope-around-in-the-dark method.

And for the future, remember that questions like this really belong in Chemistry.

Answer 11

thanks alot i get it now yeah i wasnt paying attention to what group i was in appriciate it.