Question

`"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"`?

Answer 1

Magnesium oxide.

Explanation:

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

`"Mg"("HSO"_ 3)_ (2(s)) -> "X"_ ((s)) + "H"_ 2"O"_ ((g)) + "SO"_ (2(g))`

Now, focus on balancing as many elements as you can while ignoring magnesium and the solid `"X"`.

You will end up with

`"Mg"("HSO"_ 3)_ (2(s)) -> "X"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO"_ (2(g))`

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

You will end up with

`"Mg"("HSO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO"_ (2(g))`

In order to prove that this is the identity of `"X"`, you need to use the masses of water and of sulfur dioxide produced by the reaction.

Now, you know that when `"186 g"` of magnesium bisulfite undergo decomposition, you get `"40 g"` of `"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 `"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

`"17.97 g H"_2"O" + "127.83 g SO"_2 = "145.8 g"`

This is fairly close to the known value of `"146 g"`, so you can say that the balanced chemical equation is indeed

`"Mg"("HSO"_ 3)_ (2(s)) -> "MgO"_ ((s)) + "H"_ 2"O"_ ((g)) + 2"SO"_ (2(g))`

Consequently, you can say that `"X"` is magnesium oxide, `"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 `"40 g"` to confirm the result.