# "Mg"("HSO"_3)_2 decomposes into a solid "X", "H"_2"O" vapor, and "SO"_2 gas. If "186 g" of "Mg"("HSO"_3)_2 decomposes completely and leaves behind "40 g" of solid "X", what is the identity of "X"?

Nov 26, 2017

Magnesium oxide.

#### Explanation:

Start by writing the unbalanced chemical equation that can describe this decomposition reaction.

${\text{Mg"("HSO"_ 3)_ (2(s)) -> "X"_ ((s)) + "H"_ 2"O"_ ((g)) + "SO}}_{2 \left(g\right)}$

Now, focus on balancing as many elements as you can while ignoring magnesium and the solid $\text{X}$.

You will end up with

${\text{Mg"("HSO"_ 3)_ (2(s)) -> "X"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO}}_{2 \left(g\right)}$

Notice that sulfur and hydrogen are balanced and that magnesium and oxygen are not, but that they could be depending on the identity of $\text{X}$. So a good starting point here will be to balance the equation by adding the missing atoms in place of $\text{X}$.

You will end up with

${\text{Mg"("HSO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO}}_{2 \left(g\right)}$

In order to prove that this is the identity of $\text{X}$, you need to use the masses of water and of sulfur dioxide produced by the reaction.

Now, you know that when $\text{186 g}$ of magnesium bisulfite undergo decomposition, you get $\text{40 g}$ of $\text{X}$ and

m_ ("H"_ 2"O" + "SO"_ 2) = "186 g " - " 40 g"

m_ ("H"_ 2"O" + "SO"_ 2) = "146 g"

Use the molar mass of magnesium bisulfite to convert the mass of the solid to moles.

186 color(red)(cancel(color(black)("g"))) * ("1 mole Mg"("HSO"_3)_2)/(186.45 color(red)(cancel(color(black)("g")))) = "0.9976 moles Mg"("HSO"_3)_2

According to the chemical equation that we're working with, every mole of magnesium bisulfite that undergoes decomposition produces $1$ mole of water and $2$ moles of sulfur dioxide.

In your case, the decomposition reaction will produce

0.9976 color(red)(cancel(color(black)("moles Mg"("HSO"_3)_2))) * ("1 mole H"_2"O")/(1color(red)(cancel(color(black)("mole Mg"("HSO"_3)_2)))) = "0.9976 moles H"_2"O"

and

0.9976 color(red)(cancel(color(black)("moles Mg"("HSO"_3)_2))) * "2 moles SO"_2/(1color(red)(cancel(color(black)("mole Mg"("HSO"_3)_2)))) = "1.9952 moles SO"_2

In order to check whether or not the chemical equation is correct, calculate the number of grams of water and of sulfur dioxide and compare the result to $\text{146 g}$.

0.9976 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "17.97 g"

1.9952 color(red)(cancel(color(black)("moles SO"_2))) * "64.07 g"/(1color(red)(cancel(color(black)("mole SO"_2)))) = "127.83 g"

So according to this balanced chemical equation, the reaction will produce

$\text{17.97 g H"_2"O" + "127.83 g SO"_2 = "145.8 g}$

This is fairly close to the known value of $\text{146 g}$, so you can say that the balanced chemical equation is indeed

${\text{Mg"("HSO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO}}_{2 \left(g\right)}$

Consequently, you can say that $\text{X}$ is magnesium oxide, $\text{MgO}$. To test this, use the fact that the reaction produces $1$ mole of magnesium oxide for every $1$ mole of magnesium bisulfite that undergoes decomposition and the molar mass of magnesium oxide.

0.9976 color(red)(cancel(color(black)("moles MgO"))) * "40.304 g"/(1color(red)(cancel(color(black)("mole MgO")))) = "40.21 g"

Once again, this value is fairly close to the known value of $\text{40 g}$ to confirm the result.