Question #1057d

1 Answer
Aug 25, 2017

2"Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + "O"_ (2(g)) + 4"NO"_ (2(g))2Mg(NO3)2(s)2MgO(s)+O2(g)+4NO2(g)

Explanation:

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))Mg(NO3)2(s)MgO(s)+O2(g)+NO2(g)

Now, notice that you have 22 atoms of nitrogen on the reactants' side and only 11 on the products' side.

To balance out the nitrogen, multiply the nitrogen dioxide molecules by 22. This will get you

"Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "O"_ (2(g)) + 2"NO"_ (2(g))Mg(NO3)2(s)MgO(s)+O2(g)+2NO2(g)

Next, focus on the oxygen. You have a total of 66 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))