Question

How do I convert from the mass of `"25.9 g Mg"_3"N"_2` to the mass of `"27.7 g H"_2"O"`? The reaction is magnesium nitride reacts in water to form magnesium hydroxide and ammonia.

Answer 1

I assume you know the chemical formulas already and are just unfamiliar with the math-writing notation here to communicate your reaction. So the reaction is:

`color(red)(1)"Mg"_3"N"_2(s) + color(red)(1)"H"_2"O"(l) -> color(red)(1)"Mg"("OH")_2(aq) + color(red)(1)"NH"_3(aq)`

Having `"25.9 g"` of `"Mg"_3"N"_2`, we have a specific number of `"mol"`s. We want to know the number of `"g"` of water, but we don't yet know the number of `"mol"`s of water.

These numbers of `"mol"`s are directly related to the reaction stoichiometry (the coefficients next to the formulas) above. The `1`'s are implied when nothing is done yet.

Notice how the reaction is NOT balanced yet. We need to do so to be in accordance with the law of conservation of mass (and energy).

Here is what I did:

1. You need three magnesiums on the left, so triple the `"Mg"("OH")_2(aq)`, because you can't triple single atoms in a molecule without tripling the quantity of the entire molecule.
2. You need two nitrogens on the left, so double the `"NH"_3` on the right so that you get two nitrogens on the right as well.
3. As a consequence of (1) and (2), we now have `3xx2 + 2xx3` hydrogens on the right, so we need `12` on the left, meaning that we need six waters.

`\mathbf("Mg"_3"N"_2(s) + color(green)(6)"H"_2"O"(l) -> 3"Mg"("OH")_2(aq) + 2"NH"_3(aq))`

Count up the number of magnesiums, nitrogens, oxygens, and hydrogens. It worked if they are equal on both sides.

Now, the equation is telling you that you NEED six times the number of `"mol"`s of `"H"_2"O"` as you have `"Mg"_3"N"_2` (because all magnesium nitride must react).

That means you need to convert the number of `"mol"`s of magnesium nitride to the number of `"mol"`s of water so you can translate from quantity of magnesium nitride to quantity of water.

And then you want the `"g"` of water so we have to get there from the `"mol"`s using the molar mass because that has been our goal for this problem.

So the process map for what we're doing is:

`overbrace("g Mg"_3"N"_2)^"start" stackrel("conversion 1")(->) "mols Mg"_3"N"_2 stackrel(overbrace("conversion 2")^"mol bridge")(->) "mols H"_2"O" stackrel("conversion 3")(->) overbrace("g H"_2"O")^"end"`

Using stoichiometric relationships acquired from the reaction above, we have:

`stackrel("start")overbrace("25.9" cancel("g Mg"_3"N"_2)) xx stackrel("conversion 1")overbrace((cancel("mol Mg"_3"N"_2))/("100.9494" cancel("g Mg"_3"N"_2)))xxstackrel("conversion 2")overbrace((cancel(color(green)(6) "mol H"_2"O"))/(cancel("mol Mg"_3"N"_2)))xxstackrel("conversion 3")overbrace(("18.015 g H"_2"O")/(cancel("mol H"_2"O")))`

`= stackrel("end")overbrace(color(blue)("27.7 g H"_2"O")),`

rounded to three sig figs.