Chemistry, Other
AP Chemistry
Please Provide steps thoroughly.
Consider the reaction.
Mg3N2 (s) +H2O (l) → Mg(OH)2 (s) + NH3 (g)
a) Balance the equation and name each compound.
b) How many moles of Mg(OH)2 are produced from .319 mol of magnesium nitride?
c) How many grams of water are necessary to produce 4.39 mol of
ammonia?
d) If 12.48g of magnesium hydroxide are formed, how many grams of ammonia are
produced?
-Start with 2.95g of magnesium nitride and 1.500g of water.
a) What is the limiting reactant?
b) How many grams of magnesium hydroxide can be theoretically obtained?
c) If only 2.009g of magnesium hydroxide are obtained, what is the percent yield of the
reaction?
Solution:
Mg3N2(s) + 6H2O(l) → 3 Mg(OH)2(s)+ 2 NH3(g)
Magnesium Nitride
Magnesium hydroxide
1 mol of Magnesium nitride reacts with 6 moles of water to form 3 moles of Magnesium
Hydroxide.
.319 mol of magnesium nitride x 3 mole of Magnesium hydroxide
--------------------------------------- = 0.957 mol
1 mole of Magnesium Nitride
c) How many grams of water are necessary to produce 4.39 mol of
ammonia?
From the balanced equation we know that 6 mol of water is required to produce 2
mol of ammonia.
No of moles of water produced from 4.39 mol of ammonia will be:
4.39 mol of ammonia x 6 mol of water
------------------2 mol of ammonia
= 13.17 mol of water.
Mass of water (in grams) required to produce 4.39 mol of ammonia will be =
13.17 mol of water x 18.01 g / mol of water = 237.19 g of water.
d) If 12.48g of magnesium hydroxide are formed, how many grams of ammonia
are produced?
Molecular mass of Magnesium Nitride = 100 g
Molecular mass of Magnesium Hydroxide = 58 g
100 g of Magnesium Nitride forms 174 g of Magnesium Hydroxide
Amount of Magnesium Nitride produced from 12.48 g of Magnesium hydroxide
will be = 100 x 12.48 / 174
= 7.172 g of Magnesium Nitride.
100 g of Magnesium Nitride produces 34 g of ammonia
Amount of ammonia produced from 7.172 g of Magnesium Nitride will be
= 100 x 7.172/ 34
= 21.094 g
-Start with 2.95g of magnesium nitride and 1.500g of water.
a) What is the limiting reactant?
b) How many grams of magnesium hydroxide can be theoretically obtained?
c) If only 2.009g of magnesium hydroxide are obtained, what is the percent yield of the
reaction?
Mg3N2(s) + 6H2O (l) → 3 Mg (OH)2(s)+ 2 NH3(g)
No of moles of Magnesium Nitride = 0.0295 mol
No of moles of water = 0.0833 mol
To determine the limiting reactant
0.0295 mol of Mg3N2 x 6 mol of H2O / 1 mol of Mg3N2 = 0.177 mol of H2O
0.177 mol of water is required to react with 0.0295 mol of magnesium nitride, since only
0.0833 mol of water is actually present, the amount of water is limiting so water will be
the limiting reactant.
b) How many grams of magnesium hydroxide can be theoretically obtained?
0.177 mol of H2O x 3 mol of Mg (OH) 2
----------------------- = 0.0885 mol of Mg (OH) 2
6 mol of H2O
0.0885 mol Mg(OH)2 x 58 g Mg(OH)2 / 1 mol Mg(OH)2 = 5.133 g Mg(OH)2
Percent yield = Actual yield / Theoretical yield x 100
= 2.009 / 5.133x 100
= 39.13 %
Hope the above helps!!!