limiting reactants in a chemical reaction you often have two or more reactants and until now we have...

Limiting Reactants• In a chemical reaction you often have

two or more reactants and until now we have assumed that there has been enough of each reactant for the reaction to continue to completion (both reactants totally used up).

• Limiting reactant questions give both amounts of reactants and we must determine which one runs out first (the limiting reactant).

Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first (in

this case, the H2)

Limiting Reactants

In the example below, the O2 would be the excess reagent

Steps to help you determine the limiting reactant

• Determine what information is given and what is required to calculate.

• Convert the given information to moles.• Use the mole ratio to determine the limiting

reactant and the excess reactant.• Once the limiting reactant is determined, use

it in the mole ratio calculation to determine the unknown.

• Convert the moles to the required units.

Limiting Reactant Example 1• Zn + 2 HCl

ZnCl2 + H2

• If 1.21 mol of zinc is added to 2.65 mol of HCl, which reactant is the limiting reactant?

• Mole Ratio1 Zn2 HCl

• Set up conversion using one piece of information (you chose) and cross multiply1 Zn = 1.21 Zn 2 HCl ? (1)(?) = (2)(1.21)

• Answer2.42 mol HCl

• 2.42 mol HCl is needed for the reaction. You have 2.65 mol, so HCl is the excess reactant.

• This means that Zn is the limiting reactant.

• If you had used the other piece of information to calculate the limiting reactant, this is what you would have:

1 Zn = ? Zn2 HCl 2.65

(1)(2.65) = (2)(?)• Answer

1.33 mol Zn

• 1.33 mol Zn is needed for the reaction. You only have 1.21 mol, so Zn is the limiting reactant.

• This means that HCl is the excess reactant.

• Once the limiting reactant is determined, you would use it to finish the question.

Limiting Reactant Example 2

• 2 Al + 3 CuSO4 3 Cu + Al2(SO4)3

• If 10.45 g Al reacts with 66.55 g CuSO4, what is the limiting reactant?

• Convert Al to moles10.45 g Al x 1 mol

26.981539g0.3873 mol Al

• Convert CuSO4 to moles

66.55 g CuSO4 x 1 mol159.610g

0.4170 mol CuSO4

• Mole Ratio2 Al

3 CuSO4

• Conversion and Cross Multiply

2 Al = ?3 CuSO4 0.4170

(3)(?) = (2)(0.4170)• Answer

0.2780 mol Al

• You need 0.2780 mol to complete the reaction and you have 0.3873.

• You have more than enough Al so this is the excess reactant.

• If you had used the other piece of information to calculate the limiting, this is what you would have:

• Mole Ratio2 Al3 CuSO4

• Conversion and Cross Multiply2 Al = 0.3873

3 CuSO4 ?(3)(0.3873) = (2)(?)• Answer0.5810 mol CuSO4

• You have 0.4170 mol of CuSO4 and you need 0.5810 mol, so CuSO4 is the limiting reactant.

Now that you know the limiting reactant, calculate how many mole of Cu will be formed

• Mole Ratio 3 CuSO4

3 Cu• Conversion and Cross Multiply

3 CuSO4 = 0.41703 Cu ?

(3)(0.4170) = (3)(?)• Answer

0.4170 mol Cu

Limiting Reactant Example 3

• C2H4 + 3 O2 2 CO2 + 2 H2O

• 2.80 mol ethene and 6.30 mol oxygen gas.• What is the limiting reagent?• How many grams of CO2 will be

produced?

One or the Other

• Mole Ratio1 C2H4

3 O2

• Conversion and Cross Multiply1 C2H4 = 2.803 O2 ?

(3)(2.80) = (1)(?)• Answer

8.40 mol O2

• Mole Ratio1 C2H4

3 O2

• Conversion and Cross Multiply

1 C2H4 = ? 3 O2 6.30

(3)(?) = (1)(6.30)• Answer

2.10 mol C2H4

• You have 6.30 mol of O2 and you need 8.40.

• Oxygen gas is the limiting reactant.

• You have 2.80 mol C2H4 and you only need 2.10.

• Ethene is the excess reactant.

Now that you have the limiting reactant…

• Mole Ratio 3 O2

2 CO2

• Conversion and Cross Multiply3 O2 = 6.30

2 CO2 ?(2)(6.30) = (3)(?)

• Answer4.20 mol CO2

Calculate grams

• 4.20 mol CO2 x 44.010 g1 mol

184.84 g CO2 will be formed.

(185 g)