# Question 1057d

Aug 25, 2017

$2 {\text{Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + "O"_ (2(g)) + 4"NO}}_{2 \left(g\right)}$

#### Explanation:

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

${\text{Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "O"_ (2(g)) + "NO}}_{2 \left(g\right)}$

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

${\text{Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "O"_ (2(g)) + 2"NO}}_{2 \left(g\right)}$

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 $\frac{1}{2}$. This will get you

${\text{Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + 1/2"O"_ (2(g)) + 2"NO}}_{2 \left(g\right)}$

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

${\text{Mg"("NO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + 1/2"O"_ (2(g)) + 2"NO}}_{2 \left(g\right)}$

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

$2 {\text{Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + (2 * 1/2)"O"_ (2(g)) + (2 * 2)"NO}}_{2 \left(g\right)}$

to get

$2 {\text{Mg"("NO"_ 3)_ (2(s)) -> 2"MgO"_ ((s)) + "O"_ (2(g)) + 4"NO}}_{2 \left(g\right)}$