CHEM 1364
Thursday Quiz #5
September 29, 2011
(Buckley)
1. Consider the following equation:
NH4NO3 (s) 
N2O (g) + H2O (g)
a. (1 point) Write the balanced equation.
NH4NO3 (s) 
N2O (g) +
2 H2O (g)
b. (1 point) If 7.0-mol of NH4NO3 are completely reacted, how many moles of N2O
could be formed?
7.0 mol N2O
Notice they are in a 1:1 NH4NO3:N2O ratio so it will be an equal number
OR
? mol N 2O =
7.0 mol NH 4 NO3
1 mol N 2O
×
= 7.0 mol N 2O
1
1 mol NH 4 NO3
c. (1 point) If 7.0-mol of NH4NO3 are completely reacted, how many moles of H2O
could be formed?
14 mol H2O
Notice they are in a 1:2 NH4NO3:H2O ratio so it is twice as many moles of H2O as
NH4NO3
OR
? mol H 2O =
7.0 mol NH 4 NO3
2 mol H 2O
×
= 14 mol H 2O
1
1 mol NH 4 NO3
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
2. Consider the following equation:
Mg (s)
N2 (g) 
+
Mg3N2 (s)
a. (1 point) Write the balanced equation.
3 Mg (s)
+
N2 (g) 
Mg3N2 (s)
b. (2 points) If 13.0-mol of Mg are reacted with 4.0-mol of N2, which is the limiting
reactant? Briefly explain how you determined that.
N2 is limiting since it requires 3 × 4.0 = 12.0 mol of Mg and you have 13.0 mol
Mg
Notice they are in a 3:1 Mg:N2 ratio so it is three times as many moles of Mg as
N2 so N2 is limiting
OR
Pretend Mg is limiting:
13.0 mol Mg 1 mol Mg3 N 2 13.0
? mol Mg3 N 2 =
×
=
mol Mg3 N 2
1
3 mol Mg
3
which is a little bigger than 4.0
Pretend N2 is limiting:
4.0 mol N 2 1 mol Mg3 N 2
? mol Mg3 N 2 =
×
= 4.0 mol Mg3 N 2
1
1 mol N 2
Since less Mg3N2 is formed from the 4.0 mol of N2, it is limiting.
c. (2 points) If 13.0-mol of Mg are reacted with 4.0-mol of N2, how many moles of
Mg3N2 could be formed? 4.0 mol Mg3N2 (see part b)
3. (2 points) Name both NH4NO3 and Mg3N2.
NH4NO3
ammonium nitrate
Mg3N2
magnesium nitride
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
Name ________________________
CHEM 1364
Thursday Quiz #5
September 29, 2011
(Buckley)
1. Consider the following equation:
NH4NO3 (s) 
N2O (g) + H2O (g)
a. (1 point) Write the balanced equation.
NH4NO3 (s) 
N2O (g) +
2 H2O (g)
b. (1 point) If 4.0-mol of NH4NO3 are completely reacted, how many moles of N2O
could be formed?
4.0 mol N2O
Notice they are in a 1:1 NH4NO3:N2O ratio so it will be an equal number
OR
? mol N 2O =
4.0 mol NH 4 NO3
1 mol N 2O
×
= 4.0 mol N 2O
1
1 mol NH 4 NO3
c. (1 point) If 4.0-mol of NH4NO3 are completely reacted, how many moles of H2O
could be formed?
8.0 mol H2O
Notice they are in a 1:2 NH4NO3:H2O ratio so it is twice as many moles of H2O as
NH4NO3
OR
? mol H 2O =
4.0 mol NH 4 NO3
2 mol H 2O
×
= 8.0 mol H 2O
1
1 mol NH 4 NO3
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
2. Consider the following equation:
Ba (s)
N2 (g) 
+
Ba3N2 (s)
a. (1 point) Write the balanced equation.
3 Ba (s)
+
N2 (g) 
Ba3N2 (s)
a. (2 points) If 9.0-mol of Ba are reacted with 2.0-mol of N2, which is the limiting
reactant?
N2 is limiting since it requires 3 × 2.0 = 6.0 mol of Ba and you have 9.0 mol Ba
Notice they are in a 3:1 Ba:N2 ratio so it is three times as many moles of Ba as N2
so N2 is limiting
b. OR
Pretend Ba is limiting:
9.0 mol Ba 1 mol Ba3 N 2
? mol Ba3 N 2 =
×
= 3.0 mol Ba3 N 2
1
3 mol Ba
Pretend N2 is limiting:
2.0 mol N 2 1 mol Ba3 N 2
? mol Ba3 N 2 =
×
= 2.0 mol Ba3 N 2
1
1 mol N 2
Since less Ba3N2 is formed from the 2.0 mol of N2, it is limiting.
c. (2 points) If 9.0-mol of Ba are reacted with 2.0-mol of N2, how many moles of
Ba3N2 could be formed? 2.0 mol Ba3N2 (see part b)
3. (2 points) Name both NH4NO3 and Ba3N2.
NH4NO3
Ba3N2
ammonium nitrate
barium nitride
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
Name ________________________
CHEM 1364
Thursday Quiz #5
September 29, 2011
(Buckley)
1. Consider the following equation:
NH4NO3 (s) 
N2O (g) + H2O (g)
a. (1 point) Write the balanced equation.
NH4NO3 (s) 
N2O (g) +
2 H2O (g)
b. (1 point) If 12.0-mol of NH4NO3 are completely reacted, how many moles of
N2O could be formed?
12.0 mol N2O
Notice they are in a 1:1 NH4NO3:N2O ratio so it will be an equal number
OR
? mol N 2O =
12.0 mol NH 4 NO3
1 mol N 2O
×
= 12.0 mol N 2O
1
1 mol NH 4 NO3
c. (1 point) If 12.0-mol of NH4NO3 are completely reacted, how many moles of
H2O could be formed?
24 mol H2O
Notice they are in a 1:2 NH4NO3:H2O ratio so it is twice as many moles of H2O as
NH4NO3
OR
? mol H 2O =
12.0 mol NH 4 NO3
2 mol H 2O
×
= 24 mol H 2O
1
1 mol NH 4 NO3
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
2. Consider the following equation:
Ca (s)
N2 (g) 
+
Ca3N2 (s)
a. (1 point) Write the balanced equation.
3 Ca (s)
+
N2 (g) 
Ca3N2 (s)
b. (2 points) If 8.0-mol of Ca are reacted with 2.0-mol of N2, which is the limiting
reactant?
N2 is limiting since it requires 3 × 2.0 = 6.0 mol of Ca and you have 8.0 mol Ca
Notice they are in a 3:1 Ca:N2 ratio so it is three times as many moles of Ca as N2
so N2 is limiting
OR
Pretend Ca is limiting:
8.0 mol Ca 1 mol Ca3 N 2 8.0
? mol Ca3 N 2 =
×
=
mol Ca3 N 2
1
3 mol Ca
3
which is bigger than 2.0
Pretend N2 is limiting:
2.0 mol N 2 1 mol Ca3 N 2
? mol Ca3 N 2 =
×
= 2.0 mol Ca3 N 2
1
1 mol N 2
Since less Ca3N2 is formed from the 4.0 mol of N2, it is limiting.
c. (2 points) If 8.0-mol of Ca are reacted with 2.0-mol of N2, how many moles of
Ca3N2 could be formed? 2.0 mol Ca3N2 (see part b)
3. (2 points) Name both NH4NO3 and Ca3N2.
NH4NO3
Ca3N2
ammonium nitrate
calcium nitride
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
Name ________________________
CHEM 1364
Thursday Quiz #5
September 29, 2011
(Buckley)
1. Consider the following equation:
NH4NO3 (s) 
N2O (g) + H2O (g)
a. (1 point) Write the balanced equation.
NH4NO3 (s) 
N2O (g) +
2 H2O (g)
b. (1 point) If 4.0-mol of NH4NO3 are completely reacted, how many moles of N2O
could be formed?
4.0 mol N2O
Notice they are in a 1:1 NH4NO3:N2O ratio so it will be an equal number
OR
? mol N 2O =
4.0 mol NH 4 NO3
1 mol N 2O
×
= 4.0 mol N 2O
1
1 mol NH 4 NO3
c. (1 point) If 4.0-mol of NH4NO3 are completely reacted, how many moles of H2O
could be formed?
8.0 mol H2O
Notice they are in a 1:2 NH4NO3:H2O ratio so it is twice as many moles of H2O as
NH4NO3
OR
? mol H 2O =
4.0 mol NH 4 NO3
2 mol H 2O
×
= 8.0 mol H 2O
1
1 mol NH 4 NO3
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium
2. Consider the following equation:
Sr (s)
N2 (g) 
+
Sr3N2 (s)
a. (1 point) Write the balanced equation.
3 Sr (s)
+
N2 (g) 
Sr3N2 (s)
b. (2 points) If 15.0-mol of Sr are reacted with 6.0-mol of N2, which is the limiting
reactant?
Sr is limiting since it requires 3 × 6.0 = 18.0 mol of Sr to react with 6.0 mol of N2
and you only have 15.0 mol Sr
Notice they are in a 3:1 Sr:N2 ratio so it is three times as many moles of Sr as N2
so Sr is limiting
OR
Pretend Sr is limiting:
15.0 mol Sr 1 mol Sr3 N 2
? mol Sr3 N 2 =
×
= 5.0 mol Sr3 N 2
1
3 mol Sr
Pretend N2 is limiting:
6.0 mol N 2 1 mol Sr3 N 2
? mol Sr3 N 2 =
×
= 6.0 mol Sr3 N 2
1
1 mol N 2
Since less Sr3N2 is formed from the 15.0 mol of Sr, it is limiting.
c. (2 points) If 13.0-mol of Sr are reacted with 4.0-mol of N2, how many moles of
Sr3N2 could be formed? 13/3 mol of Sr3N2 (see part b)
(I meant to say 15.0-mol of Sr to match part b, but didn’t. Sr is still limiting –
will take the answer for 13 mol or 15 mol of Sr to start with.)
3. (2 points) Name both NH4NO3 and Sr3N2.
NH4NO3
ammonium nitrate
Sr3N2
magnesium nitride
Mg  magnesium
N  nitrogen
Ca  calcium Sr  strontium
H hydrogen
O  oxygen
Ba  barium