#"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"#?

1 Answer
Nov 26, 2017

Answer:

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.