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Section 3.7

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**1. **Section 3.7 Limiting Reactants

**2. **Grilled Cheese Sandwich

**3. **LIMITING REACTANT IMPORTANCE:
Calculations of limiting reactant bring quantitative understanding to chemical reactions
These calculations are used in both General and Organic Chemistry

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

**5. **DEFINITIONS THEORETICAL YIELD
The amount of product that can be made based on the amount of the limiting reactant
ACTUAL YIELD
The amount of product actually or experimentally produced
THE PERCENT YIELD
%yield = (actual/theoretical) x 100

**6. **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.

**7. **Container 1

**8. **Before and After Reaction 1

**9. **Container 2

**10. **Before and After Reaction 2

**11. **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

**12. **METHODS USED TO DETERMINE THE LIMITING REACTANT Calculate the moles needed of each reactant and compare with the moles given
II. Divide the moles of each reactant by its stoichiometric coefficient and then compare them
Calculate the moles of product produced by each reactant and compare them

**13. **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 have
moles N2 in equation
= 3.0 mol N2
1.0 mol N2
= 3.0 (multiply all coefficients in
balanced equation by this factor)

**14. **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.

**15. **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?

**16. **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

**17. **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 mol
Ba(NO3)2 : 0.0245 mol = 0.00817 LR
3 mol

**18. **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

**19. **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?

**20. **Step 1. Convert to moles: Zn = 65 g/mol
AgNO3 = 108 + 14 + (3 x 16) = 170 g/mol
Zn = 2.0 g x 1 mol = 0.0308 mol
65 g
AgNO3 = 2.50 g x 1 mol = 0.0147 mol 170 g

**21. **Step 2. Guess limiting reagent Zn(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).

**22. ** 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.

**23. **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

**24. **Homework # 3.71-3.74 on pages 115-116

**25. **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

**26. **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)

**27. **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

**28. **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 moles
0.0607 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.

**29. **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