Section 3.7

Limiting Reactants

Grilled Cheese Sandwich

Bread + Cheese ‘Cheese Melt’

2 B + C B2C

100 bread 30 slices ? sandwiches

What is the limiting factor in our ability to make the maximum about of grilled

cheese sandwiches containing 2 slices of bread and 1 slice of cheese?

LIMITING REACTANT

• IMPORTANCE:

• Calculations of limiting reactant bring quantitative understanding to chemical reactions

• These calculations are used in both General and Organic Chemistry

DEFINITIONS• LIMITING REACTANT• Completely consumed in a chemical reaction• Determines the amount of product formed• The reactant that produces the least amount of

product

DEFINITIONS• THEORETICAL YIELD

The amount of product that can be made based on the amount of the limiting reactant

• ACTUAL YIELDThe amount of product actually or experimentally produced

• THE PERCENT YIELD%yield = (actual/theoretical) x 100

Limiting Reactants. An analogous situation occurs with chemical

reactions. Consider the reaction:

2 H2(g) + O2(g) → 2 H2O(l) 2 mol + 1 mol 2 mol

If we have exactly 2 mol of H2 and 1 mol of O2, then we can make 2 mol of water. But what if we have 4 mol of H2 and 1 mol of O2. Now we can make only 2 mol H2O with 2 mol H2 left over. In this case the O2 is the limiting reagent.

The limiting reagent is the one with nothing left over.

Container 1

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

Before and After Reaction 1

All the hydrogen and nitrogen atoms combine.

Before the reaction After the reaction

Container 2

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

Before and After Reaction 2

Before the reaction After the reaction

Multiplying an equation through by a common multiple:

We can multiply all the coefficients in a balanced equation by any multiple, and it still has the correct ratios of moles. Thus, if we have:

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) 1 mole 2 moles 1 mole + 1 mole

If we have 2 moles of Zn(s), this gives: (x 2) 2 moles 4 moles 2 moles 2 moles

or if we have 0.5 moles Zn(s) we have: (x 0.5) 0.5 moles 1 mole 0.5 moles 0.5 moles

METHODS USED TO DETERMINE THE LIMITING

REACTANTI. Calculate the moles needed of each reactant

and compare with the moles givenII. Divide the moles of each reactant by its

stoichiometric coefficient and then compare them

III. Calculate the moles of product produced by each reactant and compare them

Example I. Consider the reaction of H2 and N2 to give NH3, and assume we have 3.0 mol N2 and 6.0 mol H2.

We have the balanced equation:

N2(g)+ 3 H2(g) → 2 NH3(g) 1 mol 3 mol 2 mol

Factor = moles N2 we havemoles N2 in equation= 3.0 mol N2

1.0 mol N2

= 3.0 (multiply all coefficients in balanced equation by this factor)

Multiply all coefficients by factor (x 3):

N2(g) + 3 H2(g) → 2 NH3(g)1 mol 3 mol 2 mol

3 mol 3 x 3 = 9 mol 3 x 2 = 6 mol

Try N2 as limiting reagent:

3 mol N2 requires how many moles H2? = 3 x 3 = 9 mol

We only have 6 mol H2, so H2 is the limiting reagent.

Example II. Divide the moles of each reactant by its stoichiometric coefficient

Consider the following reaction:

2 Na3PO4(aq) + 3 Ba(NO3)2(aq) Ba3(PO4)2 + 6 NaNO3

How much Ba3(PO4)2 can be formed if we have in the solutions 3.50 g sodium phosphate and 6.40 g barium nitrate?

Step 1. Convert to moles:

First work out numbers of Moles:

Na3PO4 = 3.50 g x 1 mol = 0.0213 mol 164 g

Ba(NO3)2 = 6.40 g x 1 mol = 0.0245 mol 261 g

Step 2. Divide moles by its stoichiometric coefficient

2 Na3PO4(aq) + 3 Ba(NO3)2(aq) Ba3(PO4)2 + 6 NaNO3

Na3PO4 : 0.0213 mol = 0.01065 2 molBa(NO3)2 : 0.0245 mol = 0.00817 LR 3 mol

Example III. Calculate the amount of product produced by each reactant

1N2(g) + 3H2(g) → 2NH3(g) Given 3.0 mole 6.0 mole

3.0 mol N2 x 2 mol NH3 = 6.0 mol NH3

1 mol N2

6.0 mol H2 x 2 mol NH3 = 4.0 mol NH3 (theoretical yield)

3 mol H2

The reactant that produces the least amount of product is the L.R.H2

Practice Exercise: Zn metal (2.00 g) plus solution of AgNO3 (2.50 g)

reacts according to:

Zn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol

Which is the limiting reagent?How much Zn will be left over?

Step 1. Convert to moles:

Zn = 65 g/molAgNO3 = 108 + 14 + (3 x 16) = 170 g/mol

Zn = 2.0 g x1 mol = 0.0308 mol 65 g

AgNO3 = 2.50 g x 1 mol = 0.0147 mol 170 g

Step 2. Guess limiting reagentZn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.0308 0.0147

In this case it seems clear that AgNO3 must be the limiting reagent, because the equation says we must have 2 mols of AgNO3 for each mol of Zn(s), but in fact we have more moles of Zn(s).

We can check this by dividing the moles of each reactant by their coefficients AgNO3 = 0.0147/2 = 0.00735

Zn = 0.0308/1 = 0.0308

Zn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.0308 0.00735

We in fact have 0.0305 mol of Zn, which is more than the 0.00735 mol of AgNO3, so AgNO3 is clearly the limiting reactant.

How much Zn is left over?• Use the limiting reactant to determine this: 0.0147 mol AgNO3 x 1 mol Zn x 65 g Zn

2 mol AgNO3 1 mol Zn

= 0.47775 g Zn• Subtract this from the amount of Zn available:

2.00 g Zn - 0.4775g Zn = 1.52 g Zn in excess

Homework• # 3.71-3.74 on pages 115-116

Percent Yield:Theoretical yields:The quantity of product that forms if all of the limiting reagent reacts is called the theoretical yield. Usually, we obtain less than this, which is known as the actual yield.

Percent yield = actual yield x 100 Theoretical yield

Problem:10.4 g of Ba(OH)2 was reacted with an excess of Na2SO4 to give a precipitate of BaSO4. If the reaction actually yielded 11.2 g of BaSO4, what is a) the theoretical yield of BaSO4 and b) what is the percentage yield of BaSO4?The balanced equation for the reaction is:

Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)

Step 1. Convert to moles:Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)1 mole 1 mole 1 mole 2 moles

Moles Ba(OH)2:

Mol. Mass Ba(OH)2 = 137.3 + 2 x (16.0 + 1.0) = 171.3 g/mol

Moles = 10.4 g x 1 mol = 0.0607 moles 171.3 g

Step 2. Work out how much BaSO4 will be formed:

Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)1 mole 1 mole 1 mole 2 moles0.0607 moles 0.0607 moles

When it says that one reagent is in excess, that means we do not have to worry about that reagent, and the other one is the limiting reagent, in this case the BaSO4. We see that 1 mole of Ba(OH)2 will produce 1 mole of BaSO4. Our factor is thus 0.0607, and we will get 0.0607 moles of BaSO4.

Convert actual yield to percentage yield:

Percent yield = actual yield x 100 % Theoretical yield

= 11.2 g x 100 %14.29 g

= 78.4 % yield