Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation for the reaction.

2ErF3 +3Mg = 3MgF2 + 2Er

Explanation:

This is a single replacement equation where there are 2 metals. The bonds are broken and new bonds are formed again by Mg and F.

Er has a +3 charge and F has a -1 charge. You switch it around and you get ErF3. Then you add the second reactant, Mg. The product is MgF as stated, and Mg has a charge of +2 and F has -1. You switch it again and you get MgF2. Then the second product Er is there.

Now we have

ErF3+Mg=MgF2+Er

So we balance the equation because of the law of conservation of mass.

Make F equal, so we add the coefficents 2 and 3

2ErF3+Mg=3MgF2+Er

And now Mg and Er need balancing so

2ErF3+3Mg=3MgF2+2Er

Hope this helped

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explanation: