I am trying to Balance this Equation and having troubles.

I am supposed to put the co efficients in the spaces.

___MgF2 + Li2CO3=___MgCO3+___LiF


Here is also another one that I did but want to make sure I did it right?

___LiCl+___Br2=___LiBr+___Cl2

THANKS FOR YOUR HELP AHEAD OF TIME!! Very much appreciation~~~

The object is to have the same amount of the same types of atoms on both side of the equations.

MgF_2 + Li_2CO_3 = MgCO_3 + 2LiF

2LiCl + Br_2 = 2LiBr + Cl_2

My teacher has us make a t chart with the number of atoms on each side, then we add coefficients and cross out and calculate the new number of atoms until they are equal.

Excellent method, but I do most of it in my head. I've been working with these types of equations for years.

So do I but some of the brightest students have to use this method.

Oh, I understand.

So what should go in the four spaces in order should be

2, 1, 2, 1 ?

Here is one more, I would like to know if I did this one right also?

___PbBr2 +___ HCl=___HBr+___PbCl2

1 2 2 2??

1 \, PbBr_2 + 2\, HCl \rightarrow 2\, HBr + 2\, PbCl_2\,\,\,???

Almost, but it's not quite right. You have one lead atom on the left and two chlorine atoms on the left. On the right, you have two lead atoms and four chlorine atoms.

[math]1PbBr_2+HCl/rightarrow2HBr+1PbCl_2

Sorry,
1PbBr_2+HCl/rightarrow2HBr+1PbCl_2

1221

Here is one more, I would like to know if I did this one right also?

___PbBr2 +___ HCl=___HBr+___PbCl2

1 2 2 2????????????????

Ok, in your head, look at Pb first. There's one on both sides, so good.
Then, you see Br, there are two of them on the left and only one on the right. That means that you have to put '2' in from of HBr.

Doing so doubles the H. Since there's only one on the left hand side, double the HCl. That means you also have 2 Cl, but you already have 2 Cl on the left!

Ok, you saw everything. Time to check.
1 Pb on each side. Good
2 Br on each side. Good
2 H on each side. Good
2 Cl on each side. Good.

That's how I do, a sort of 'chaining' with the elements I'm dealing with, trying to fix one as soon as possible. That works for the easy equations. Other, more difficult require perseverance, and sometimes the use of half equations and fractions.