Question #4ceea
1 Answer
Here's what I got.
Explanation:
Interestingly enough, adding water to magnesium nitride,
Once the ammonia is evolved, the magnesium hydroxide is heated and decomposes to form magnesium oxide,
So, you can say that you have
#"Mg"_3"N"_text(2(s]) + color(red)(6)"H"_2"O"_text((l]) -> color(blue)(3)"Mg"("OH")_text(2(s]) + 2"NH"_text(3(g]) uarr#
followed by
#"Mg"("OH")_text(2(s]) stackrel(color(purple)("heat")color(white)(xx))(->) "MgO"_text((s]) + "H"_2"O"_text((g]) uarr#
So, your strategy here will be to use the given masses of magnesium nitride and water and the
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
#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
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_(MgO) = color(green)("0.114 moles MgO")#
#m_(MgO) = color(green)("4.58 g MgO")#