Limiting Reactant Class Work Ans
2N2 H4 + N2O4 ® 3N 2 + 4H 2O
1. Provided you begin with 50.00 moles of N2H4 and 40.00 moles of N2O4 :
a. What is the limiting reactant for this problem?
Available
50.00 moles N2H4
40.00 moles N2O4
4H 2O
4H 2O
50.00
= 100.0 mol H2O
40.00
= 160.0 mol H2O
2N 2 H 4
1N 2O4
100.0 < 160.0
100.0 mol of H2O can be produced
50.00 mol of N2H4 will be consumed
Ignore the 40 and 160
All N2H4 is consumed so N2H4 is limiting
b. Find the actual moles of H2O produced.
100.0 mol H2O produced
c. How many moles of the non-limiting reactant remain at the
completion of the reaction?
Either the numbers known can be used to start this part
N 2O4
1N 2O4
50.00
= 25.00 mol N2O4
100.0
= 25.00 mol N2O4
2N 2 H 4
4H 2O
Started with 40.00 and 25.00 is consumed
40.00 – 25.00 = 15.00 mol N2O4 remain
2N2 H4 + N2O4 ® 3N 2 + 4H 2O
2. If you begin with 400.0 g of N2H4 and 900.0 g of N2O4 :
a. Find the mass of N2 produced.
Available
12.48 mol N2H4
9.782 mol N2O4
12.48
3N 2
3N 2
= 18.72 mol N2
9.782
= 29.34 mol N2H4
2N 2 H 4
1N 2O4
so 12.48 mol N2H4 is consumed and 18.72 mol N2 is produced
18.72 (28.014) = 524.5 g of N2
b. What is the limiting reactant for this problem? N2H4
c. How many moles of the excess reactant remain at the completion of
the reaction?
Since N2H4 is limiting, look for N2O4. Calculate how much is consumed.
1N 2O4
1N 2O4
12.48
= 6.241 mol N2O4 18.72
= 6.241 mol N2O4
2N 2 H 4
3N 2
Subtract that from initial N2O4
9.782 – 6.241 = 3.541 mol N2O4 remain
d. How many moles of H2O will be produced in this reaction?
4H 2O
12.48
= 24.96 mol H2O
2N 2 H 4
2Al + 3Cl2 ® 2AlCl3
3. Aluminum metal and Chlorine gas react to form Aluminum Chloride, often
used as a catalyst for many reactions performed by the chemical industry. A
Reactor initially contains 200.0 g of Al metal and 1125 g of Cl2 gas:
a. Find the actual mass of AlCl3 produced.
7.412 mol Al
15.866 mol Cl2
2AlCl3
2AlCl3
7.412
= 7.412 mol AlCl3
15.866
= 10.58 mol AlCl3
3Cl2
2Al
So 7.412 mol Al consumed and 7.412 mol AlCl3 produced
7.412 (133.341) = 988.4 g of AlCl3
b. What is the limiting reactant for this problem?
Al
c. How many moles of the limiting reactant remain at the completion of
the reaction?
None
__ C3H8 + 5O2 ® 3CO2 + 4H 2O
4. Propane: C3H8, is used as a fuel in the presence of oxygen. Within an
enclosed container only 96.1 g of C3H 8 and 325 g of O 2 are present.
a. Which of the two reactants will cause the fire to go out first?
96.1
= 2.18 mol C3 H 8
44.096
2.18 mol C3 H 8
10.2 mol O2
325
= 10.2 mol O2
31.998
5O2
= 10.9 mol O2
1C3 H 8
1C3 H 8
= 2.04 mol C3 H 8
5O2
Starting materials
not possible, not enough O2
this is what is actually consumed
Since 10.2 mol O2 is available and 10.2 mol of O2 is consumed the oxygen will run
out first. O2 is the limiting reactant.
b. How many grams of the non-limiting reactant will remain after the
completion of the reaction?
Had 2.18 mol C3 H 8 to start with and used 2.04 mol C3 H 8
2.18 – 2.04 = 0.14 mol C3 H 8 left =
6.2 g of C3H8
c. How many moles of CO2 will be produced?
Here you can use several different numbers
3CO2
10.2 mol O2
= 6.12 mol CO2
5O2
2.04 mol C3 H 8
3CO2
= 6.12 mol CO2
1C3 H 8
d. How many moles of H2O will be produced?
10.2 mol O2
4H 2O
= 8.16 mol H 2O
5O2
2.04 mol C3 H 8
4H 2O
= 8.16 mol H 2O
1C3 H 8
2AlCl3
7.418 2Al