1. Chemical Equations Stoichiometry V
Limiting Reagents
Percent Yield
Actual Yield
Theoretical Yield
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2. Limiting Reagent
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3. What is the limiting reagent?
Reactant or reagent that is completely consumed
Determines the quantities of products that form
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4. Problem Solving Strategy
Do two or more calculations in parallel (make two or more assumptions)
Choose the correct one
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5. Example
If 1.00 kg each of PCl3, Cl2 and P4O10 are allowed to react, what mass of POCl3, in kg, will be formed?
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6. Step 1 Write a balanced equation
6PCl3 + 6 Cl2 + P4O > 10 POCl3
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7. Step 2
Assume each reactant is used up. The smallest amount is the limiting reagent.
Mass given in kg? We can use kmol instead of mol.
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8. 6PCl3 + 6 Cl2 + P4O10 --- > 10 POCl3
Reagent 1. PCl3
6PCl3 + 6 Cl2 + P4O > 10 POCl3
1.00 kg PCl3 x 1 kmol PCl3 x 10 kmol POCl3
137 kg PCl kmol PCl3
= kmol POCl3
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9. 6PCl3 + 6 Cl2 + P4O10 --- > 10 POCl3
Reagent 2. Cl2
6PCl3 + 6 Cl2 + P4O > 10 POCl3
1.00 kg Cl2 x 1 kmol Cl2 x 10 kmol POCl3
71.0 kg Cl kmol Cl2
= kmol POCl3
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10. 6PCl3 + 6 Cl2 + P4O10 --- > 10 POCl3
Reagent 3. P4O10
6PCl3 + 6 Cl2 + P4O > 10 POCl3
1.00 kg P4O10 x 1 kmol P4O10 x 10 kmol POCl3
284 kg P4O kmol P4O10
= kmol POCl3
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11. Smallest amount is the limiting reagent *PCl3 = 0.0121 kmol POCl3
P4O10 = kmol POCl3
* Limiting reagent
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12. Step 3
Determine the amount of product produced using the amount produced by the limiting reagent.
kmol POCl3 x kg POCl3
1 kmol POCl3
= 1.86 kg POCl3
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13. Determining the Amount in Excess
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14. Total amount of reactant – amount of reactant that reacted
Determine mass of each reagent using the given mass of the limiting reagent
Subtract:
Total amount of reactant – amount of reactant that reacted
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15. Reagent 1. PCl3 Amount in Excess 1.00 kg – 1.00 kg = 0.00 kg PCl3
Started with 1.00 kg PCl3
Amount Reacted: 1.00 kg
Amount in Excess
1.00 kg – 1.00 kg = 0.00 kg PCl3
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16. Reagent 2. Cl2 Amount in Excess
1.00 kg PCl3 x 1 kmol PCl3 x 6 kmol Cl2 x 71 kg Cl2
137 kg PCl kmol PCl kmol Cl2
= kg Cl (Amount Reacted)
Amount in Excess
1.00 kg – kg = kg = 0.48 kg Cl2
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17. = 0.345 kg P4O10 (Amount Reacted) Amount in Excess
Reagent 3. P4O10
1.00 kg PCl3 x 1 kmol PCl3 x 1 kmol P4O10 x kg P4O10
137 kg PCl kmol PCl kmol P4O10
= kg P4O10 (Amount Reacted)
Amount in Excess
1.00 kg – kg = 0.66 kg P4O10
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18. You can also start with 0.0121 kmol POCl3 and get the same answer.
kmol POCl3 x kmol Cl2 x kg Cl2
10 kmol POCl kmol Cl2
= kg Cl2
1.00 kg – kg = kg Cl2 = 0.49 kg Cl2
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19. Your Turn
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20. If 12.2 g H2 and 154 g of O2 are allowed to react, which gas and what mass of that gas remains after the reaction?
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21. Balanced Equation 2 H2(g) + O2(g) ----- > 2 H2O(l) 12/26/2018
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22. 2 H2(g) + O2(g) ----- > 2 H2O(l)
Reagent 1. H2
2 H2(g) + O2(g) > 2 H2O(l)
12.2 g H2 x 1 mol H2 x 2 mol H2O
2.02 g H mol H2=
= 6.10 mol H2O
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23. 2 H2(g) + O2(g) ----- > 2 H2O(l)
Reagent 2. O2
2 H2(g) + O2(g) > 2 H2O(l)
154 g O2 x 1 mol O2 x 2 mol H2O
32.0 g O mol O2
= 9.63 mol H2O
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24. H2 = 6.10 mol H2O (limiting reagent) O2 = 9.63 mol H2O
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25. Mass of gas in Excess Reagent 1. H2 H 2 reacted = 12.2 g Amount left
12.2 g – 12.2 g = 0.00 g
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26. Reagent 2. O2 12.2 g H2 x 1 mol H2 x 1 mol O2 x 32.0 g O2
2.02 g H mol H mol O2
= 97.1 g O2
Amount in Excess:
154 g – 97.1 g = 56.9 g O2 = 57 g O2
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27. Percent, Actual and Theoretical Yield
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28. Reasons why yield isn’t 100%
Reactants or products leak out, especially when they are gases.
The reactants are not 100% pure
Some product is lost when it is purified.
The product decomposes back into reactants.
The products react to form different substances.
The reaction occurs very slowly.
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29. Formula Percent yield = actual yield x 100% theoretical yield
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30. Example 1
What is the percent yield, if the quantity of reactants is sufficient to produce 0.86 g of Cl2O but only g is obtained?
% yield = g x 100% = 83%
0.86 g
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31. Example 2
Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g) to produce methanol kg of CH3OH is actually produced.
Calculate the theoretical yield of methanol.
What is the percent yield of methanol?
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32. 2H2 + CO(g) --- > CH3OH a. 68.5 kg CO x 1 kmol CO x 1 kmol CH3OH
28.0 kg CO mol CO
= 2.45 kmol CH3OH
8.60 kg H2 x 1 kmol H2 x 1 kmol CH3OH
2.02 kg H kmol H2
= kmol CH3OH
H2 is the limiting reactant.
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33. % Yield % yield = 52.0 % Theoretical yield
2.14 kmol CH3OH x kg CH3OH = 68.6 kg CH3OH
1 kmol CH3OH
% yield = 35.7 kg x 100 %
68.6 kg
% yield = 52.0 %
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34. Homework Wkst. Limiting Reactants I
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35. 1.72 mol ZnS x 2 mol ZnO = 1.72 mol ZnO 2 mol ZnS
1) 2 ZnS + 3 O2 --- > 2 ZnO SO2
1.72 mol ZnS x 2 mol ZnO = 1.72 mol ZnO
2 mol ZnS
3.04 mol O2 x 2 mol ZnO = 2.03 mol ZnO
3 mol O2
ZnS is the limiting reactant
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36. 2a) Al + 3O2 --- > 2Al2O3
0.32 mol Al x 2 mol Al2O3 = mol Al2O3
4 mol Al
0.26 mol O2 x 2 mol Al2O3 = 0.17 mol Al2O3
3 mol O2
Al is the limiting reactant.
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37. 2b) Al + 3 O2 --- > 2Al2O3
6.38 x 10-3 mol O2 x 2 mol Al2O3= mol Al2O3
3 mol O2
9.15 x 10-3 mol Al x 2 mol Al2O3 = mol Al2O3
4 mol Al
O2 is the limiting reactant
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38. 2c) Al + 3 O2 --- > 2Al2O3
3.17 g Al x 1 mol Al x 2 mol Al2O3 = mol Al2O3
27.0 g Al 4 mol Al
2.55 g O2 x 1 mol O2 x 2 mol Al2O3= mol Al2O3
32.0 g O2 3 mol O2
O2 is the limiting reactant.
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39. CuS is the limiting reactant.
3a. 2 CuS 3 O2 -- > 2 CuO + 2 SO2
100 g CuS = 1.04 mol CuO
56 g O2 = 1.17 mol CuO
CuS is the limiting reactant.
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40. CuS is the limiting reactant.
3b. 2 CuS + 3O2 -- > 2 CuO + 2 SO2
18.7 g CuS = 15.6 g CuO
12.0 g O2 = 20.0 g CuO
CuS is the limiting reactant.
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41. 4. 3 CuSO4 + 2 Fe -- > 3 Cu + Fe2(SO4)3
0.092 mol Fe = 0.14 mol Cu
O.158 mol CuSO4 = mol Cu
0.14 moles of Cu will form.
Fe is the limiting reactant
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42. 5. BaCO3 + 2 HNO3 -- > Ba(NO3)2 + CO2 + H2O
55 g BaCO3 = 73 g Ba(NO3)2
26 g HNO3 = 54 g Ba(NO3)2
HNO3 is the limiting reactant.
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43. 6a) MgI2 + Br2 -- > MgBr2 + I2
560 g MgI2 = 2.0 mol I2
360 g Br2 = 2.3 mol I2
Br2 is the excess reactant.
2.0 mol I2 x 1 mol Br2 x 160 g Br2 =320gBr2
1 mol I mol Br2
360 g Br2 – 320 g Br2 = 40. g Br2
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44. 6b) MgI2 + Br2 -- > MgBr2 + I2
2.0 mol I2 x g I2 = 5.1 x 102 g I2
1 mol I2
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45. 7a) 2 AgNO3 + Ni -- > 2 Ag + Ni(NO3)2
22.9 g Ni = mol Ni(NO3)2
112 g AgNO3 = mol Ni(NO3)2
Ni is in excess.
AgNO3 is the limiting reactant..
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46. 7b) 2 AgNO3 + Ni -- > 2 Ag + Ni(NO3)2
0.390 mol Ni(NO3)2 = 71.4 gNi(NO3)2
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47. 8. CS2 + 3O2 -- > 2SO2 + CO2 1.60 mol CS2 = 3.20 mol SO2
5.60 mol O2 = 3.73 mol SO2
O2 is in excess
1.60 mol CS2 x 3 mol O2 = 4.80 mol O2
1 mol CS2
5.60 mol O2 – 4.80 mol O2 = 0.80 mol O2
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48. Classwork I Limiting Reactants II#1
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49. 1a. HgCl2 + 2 NH3 --- > Hg(NH2)Cl + NH4Cl
0.91 g HgCl2 x 1 mol HgCl2 x 1 mol Hg(NH2)Cl x 252 g Hg(NH2)Cl
272 g HgCl mol HgCl mol Hg(NH2)Cl
= 0.84 g
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50. 1b. HgCl2 + 2 NH3 --- > Hg(NH2)Cl + NH4Cl
0.15 g NH3 x 1 mol NH3 x 1 mol Hg(NH2)Cl x 252 g Hg(NH2)Cl
17.0 g NH mol NH mol Hg(NH2)Cl
= 1.1 g Hg(NH2)Cl
0.91 g HgCl2 x 1 mol HgCl2 x 1 mol Hg(NH2)Cl x 252 g Hg(NH2)Cl
272 g HgCl mol HgCl mol Hg(NH2)Cl
= 0.84 g Hg(NH2)Cl (limiting reactant)
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51. Lab Report Cu-AgNO3 Reaction
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52. Outline Name, Date, Period # in top right hand corner Title (centered)
Purpose
Hypothesis
Materials
Observations
Data Table
Calculations
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53. Calculations (Labeled)
Change in mass of the copper wire
Number of moles of copper reacted
Mass of silver obtained
Number of moles of silver
Ag/Cu Ratio
Number of moles of silver nitrate used in the experiment
Ag/AgNO3 Ratio
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54. Atoms of copper removed from the wire Atoms of silver metal formed
Word Equation
Atoms of copper removed from the wire
Atoms of silver metal formed
Limiting Reactant
Amount of substance in excess
Theoretical yield of Silver
Percent Yield
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