# Question 4ceea

Dec 24, 2015

Here's what I got.

#### Explanation:

Interestingly enough, adding water to magnesium nitride, ${\text{Mg"_3"N}}_{2}$, will not produce magnesium oxide, it will actually produce magnesium hydroxide, "Mg"("OH")_2, and ammonia, ${\text{NH}}_{3}$.

Once the ammonia is evolved, the magnesium hydroxide is heated and decomposes to form magnesium oxide, $\text{MgO}$, and water.

So, you can say that you have

${\text{Mg"_3"N"_text(2(s]) + color(red)(6)"H"_2"O"_text((l]) -> color(blue)(3)"Mg"("OH")_text(2(s]) + 2"NH}}_{\textrm{3 \left(g\right]}} \uparrow$

followed by

${\text{Mg"("OH")_text(2(s]) stackrel(color(purple)("heat")color(white)(xx))(->) "MgO"_text((s]) + "H"_2"O}}_{\textrm{\left(g\right]}} \uparrow$

So, your strategy here will be to use the given masses of magnesium nitride and water and the $1 : \textcolor{red}{6}$ mole ratio that exists between these two reactants to determine whether or not you're dealing with a limiting reagent.

So, determine how many moles of each reactant you have by using the molar masses of the two compounds

3.82 color(red)(cancel(color(black)("g"))) * ("1 mole Mg"_3"N"_2)/(100.93 color(red)(cancel(color(black)("g")))) = "0.03785 moles Mg"_3"N"_2

7.73 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015 color(red)(cancel(color(black)("g")))) = "0.4291 moles H"_2"O"

So, how many moles of water would you need in order for all the moles of magnesium nitride to react?

0.03785 color(red)(cancel(color(black)("moles Mg"_3"N"_2))) * (color(red)(6)" moles H"_2"O")/(1color(red)(cancel(color(black)("mole Mg"_3"N"_2)))) = "0.2271 moles H"_2"O"

Since you have more moles of water than would be needed, you can say that magnesium nitride will act as a limiting reagent.

Now, notice that $1 : \textcolor{b l u e}{3}$ mole ratio that exists between magnesium nitride and magnesium hydroxide. This tells you that the first reaction will produce - remember, all the magnesium nitride reacts!

0.03785 color(red)(cancel(color(black)("moles Mg"_3"N"_2))) * (color(blue)(3)" moles Mg"("OH")_2)/(1color(red)(cancel(color(black)("mole Mg"_3"N"_2)))) = "0.1136 moles Mg"("OH")_2

Finally, the $1 : 1$ mole ratio that exists between magnesium hydroxide and magnesium oxide tells you that the decomposition reaction reaction will produce $0.1136$ moles $\text{MgO}$.

Use magnesium oxide's molar mass to determine how many grams would contain this many moles

0.1136 color(red)(cancel(color(black)("moles MgO"))) * "40.3 g"/(1color(red)(cancel(color(black)("mole MgO")))) = "4.5781 g MgO"#

Rounded to three sig figs, the answers will be

${n}_{M g O} = \textcolor{g r e e n}{\text{0.114 moles MgO}}$

${m}_{M g O} = \textcolor{g r e e n}{\text{4.58 g MgO}}$