Question #1057d

Start by writing the unbalanced chemical equation that describes the decomposition of magnesium nitrate

#"Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "O"_ (2(g)) + "NO"_ (2(g))#

Now, notice that you have #2# atoms of nitrogen on the reactants' side and only #1# on the products' side.

To balance out the nitrogen, multiply the nitrogen dioxide molecules by #2#. This will get you

#"Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "O"_ (2(g)) + 2"NO"_ (2(g))#

Next, focus on the oxygen. You have a total of #6# atoms of oxygen on the reactants' side and

#overbrace("1 O")^(color(blue)("from MgO")) + overbrace("2 O")^(color(blue)("from O"_2)) + overbrace("4 O")^(color(blue)("from 2 NO"_2)) = "7 O"#

on the products' side. To balance out the oxygen, multiply the oxygen molecule by #1/2#. This will get you

#"Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + 1/2"O"_ (2(g)) + 2"NO"_ (2(g))#

The products' side will now have

#overbrace("1 O")^(color(blue)("from MgO")) + overbrace("1 O")^(color(blue)("from"color(white)(.)1/2"O"_2)) + overbrace("4 O")^(color(blue)("from 2 NO"_2)) = "6 O"#

Since you have #1# atom of magnesium on both sides of the equation, you can say that the balanced chemical equation that describes the decomposition of magnesium nitrate looks like this

#"Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + 1/2"O"_ (2(g)) + 2"NO"_ (2(g))#

If you want, you can get rid of the Fractional coefficient by multiplying all the chemical species involved in the reaction by #2#

#2"Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + (2 * 1/2)"O"_ (2(g)) + (2 * 2)"NO"_ (2(g))#

to get

#2"Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + "O"_ (2(g)) + 4"NO"_ (2(g))#