Limiting Reagents 

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Problem One


1. For the balanced equation FeS+2HCl=>FeCl2+H2S, 
what would the limiting reagent be if 67.9 grams of FeS were reacted with 87.4 grams of HCl?

Step One: Define the ratio of the reagents
1:2 ratio

Step Two: Find the molar mass of each reagent
Fe * 1 = 55 * 1 = 55                                     H * 1 = 1 * 1 = 1
S * 1 = 32 * 1 = 32                                       Cl * 1 = 35 * 1 = 35

55 + 32 = 87 grams/mole                            1 + 35 = 36 grams/mole

Step Three: Convert grams to moles
67.9 grams = 0.78 moles                              87.4 grams = 2.42 moles
87 g/m                                                           36 g/m

0.78 * 1 = 0.78 moles                                 2.42 * 2 = 4.84 moles

Step Four: Identify the limiting reagent
Fes is the limiting reagent


Problem Two

2. For the balanced equation 2H2S+3O2=>2H2O+2SO2, 
what would the limiting reagent be if 21.1 grams of H2S were reacted with 40.4 grams of O2?

Step One: Define the ratio of the reagents
2:3 ratio

Step Two: Find the molar mass of each reagent
H * 2 = 1 * 2 = 2                                    O * 2 = 16 * 2 = 32 grams/mole
S * 1 = 32 * 1 = 32

2 + 32 = 34 grams/mole

Step Three: Convert grams to moles
21.1 grams = 0.62 moles                        40.4 grams = 1.26 moles
34 g/m                                                      32 g/m

0.62 * 2 = 1.24 moles                           1.26 * 3 = 3.78 moles

Step Four: Identify the limiting reagent
H2S is the limiting reagent


Problem Three
3. For the balanced equation 4C2H3O2F+7O2=>8CO2+6H2O+2F2, 
what would the limiting reagent be if 69.7 grams of C2H3O2F were reacted with 35.9 grams of O2?

Step One: Define the ratio of the reagents
4:7 ratio

Step Two: Find the molar mass of each reagent
C * 2 = 12 * 2 = 24                                    O * 2 = 16 * 2 = 32 grams/mole
H * 3 = 1 * 3 = 3
O * 2 = 16 * 2 = 32
F * 1 = 19 * 1 = 19

24 + 3 + 32 + 19 = 78 grams/mole

Step Three: Convert grams to moles
69.7 grams = 0.89 moles                           35.9 grams = 1.12 moles
78 g/m                                                         32 g/m

0.89 * 4 = 3.56 moles                             1.12 * 7 = 7.84 moles

Step Four: Identify the limiting reagent
O2 is the limiting reagent


Problem Four
4. For the balanced equation 4C2H3Cl+7O2=>8CO+6H2O+2Cl2, 
what would be the limiting reagent if 22.6 grams of C2H3Cl were reacted with 30.0 grams of O2?

Step One: Define the ratio of the reagents
4:7 ratio

Step Two: Find the molar mass of each reagent
C * 2 = 12 * 2 = 32                                 O * 2 = 16 * 2 = 32 grams/mole
H * 3 = 1 * 3 = 3
Cl * 1 = 35 * 1 = 35

32 + 3 + 35 = 70 grams/mole

Step Three: Convert grams to moles
22.6 grams = 0.32 moles                      30.0 grams = 0.95 moles
70 g/m                                                    32 g/m

Step Four: Identify the limiting reagent
C2H3Cl is the limiting reagent


Problem Five
5. For the balanced equation 2C3H6+9O2=>6CO2+6H2O, 
what would the limiting reagent be if 49.7 grams of C3H6 were reacted with 91.8 grams of O2?

Step One: Define the ratio of the reagents2:9 ratio
Step Two: Find the molar mass of each reagent
C * 3 = 12 * 3 = 36                                 O * 2 = 12 * 2 = 32 grams/mole
H * 6 = 1 * 6 = 6

36 + 6 = 42 grams/mole

Step Three: Convert grams to moles
49.7 grams = 1.18 moles                      91.8 grams = 2.86 moles
42 g/m                                                    32 g/m

Step Four: Identify the limiting reagent
O2 is the limiting reagent


Problem Six
6. For the balanced equation CH3COF+H2O=>CH3COOH+HF, 
what would the limiting reagent be if 88.9 grams of CH3COF were reacted with 19.3 grams of H2O?

Step One: Define the ratio of the reagents
1:1 ratio

Step Two: Find the molar mass of each reagent
C * 1 = 12 * 1 = 12                                       H * 2 = 1 * 2 = 2
H * 3 = 1 * 3 = 3                                           O * 1 = 16 * 1 = 16
C * 1 = 12 * 1 = 12
O * 1 = 16 * 1 = 16
F * 1 = 19 * 1 = 19

12 + 3 + 12 + 16 + 19 = 62 grams/mole      2 + 16 = 18 grams/mole

Step Three: Convert grams to moles
88.9 grams = 1.43 moles                           19.3 grams = 1.07 moles
62 g/m                                                         18 g/m

Step Four: Identify the limiting reagent
H2O is the limiting reagent


Problem Seven
7. For the balanced equation N2O5+H2O=>2HNO3, 
what would the limiting reagent be if 36.9 grams of N2O5 were reacted with 4.32 grams of H2O?

Step One: Define the ratio of the reagents
1:1 ratio

Step Two: Find the molar mass of each reagent
N * 2 = 14 * 2 = 28                                H * 2 = 1 * 2 = 2
O * 5 = 16 * 5 = 80                                O * 1 = 16 * 1 = 16

28 + 80 = 108 grams/mole                   2 + 16 = 18 grams/mole

Step Three: Convert grams to moles
36.9 grams = 0.34 moles                     4.32 grams = 0.24
108 g/m                                               18 g/m

Step Four: Identify the limiting reagent
H2O is the limiting reagent


Problem Eight
8. For the balanced equation C6H6S+6O2=>6CO+3H2O+SO3, 
what would the limiting reagent be if 67.8 grams of C6H6S were reacted with 206 grams of O2?

Step One: Define the ratio of the reagents
1:6 ratio

Step Two: Find the molar mass of each reagent
C * 6 = 12 * 6 = 72                                O * 2 = 16 * 2 = 32 grams/mole
H * 6 = 1 * 6 = 6
S * 1 = 32 * 1 = 32

72 + 6 + 32 = 110 grams/mole

Step Three: Convert grams to moles
67.8 grams = 0.616 moles                 206 grams = 6.43 moles
110 g/m                                              32 g/m

Step Four: Identify the limiting reagent
C6H6S is the limiting reagent


Problem Nine
9. For the balanced equation SiO2+3C=>SiC+2CO, 
what would the limiting reagent be if 7.27 grams of SiO2 were reacted with 2.67 grams of C?

Step One: Define the ratio of the reagents
1:3 ratio

Step Two: Find the molar mass of each reagent
Si * 1 = 28 * 1 = 28                                C * 1 = 12 * 1 = 12 grams/mole
O * 2 = 16 * 2 = 32

28 + 32 = 60 grams/mole                                        

Step Three: Convert grams to moles
7.27 grams = 0.121 moles                   2.67 grams = 0.222 moles
60 g/m                                                 12 g/m

Step Four: Identify the limiting reagent
C is the limiting reagent


Problem Ten
10. For the balanced equation 2C2H3O2Cl+O2=>4CO+2H2O+2HCl, 
what would the limiting reagent be if 28.0 grams of C2H3O2Cl were reacted with 3.69 grams of O2?

Step One: Define the ratio of the reagents
2:1 ratio

Step Two: Find the molar mass of each reagent
C * 2 = 12 * 2 = 24                                O * 2 = 16 * 2 = 32 grams/mole
H * 3 = 1 * 3 = 3
O * 2 = 16 * 2 = 32
Cl * 1 = 35 * 1 = 35

24 + 3 + 32+ 35 = 94 grams/mole

Step Three: Convert grams to moles
28 grams = 0.29 moles                       3.69 grams = 0.11 moles
94 g/m                                                 32 g/m

Step Four: Identify the limiting reagent
O2 is the limiting reagent