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

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.