Chemical Reaction Stoichiometry. Practical matters in chemical reaction stoichiometry. Haber process to produce ammonia. production of ammonia from Calcium cyanamide. Manganese dioxide to produce chlorine.

1. Reaction Stoichiometry
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. Practical Matters in Reaction Stoichiometry
Percent yield represents the ratio between what is experimentally
obtained and what is theoretically calculated, multiplied by 100%.
% 𝑦𝑖𝑒𝑙𝑑 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥 100
how much water is produced when 12.0 g of glucose (C6H12O6) is burned with enough
oxygen.
C6H12O6 + 6O2 → 6CO2 + 6H2O
12.0 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑥
1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
180.0 𝑔
𝑥
6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑟
1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
𝑥
18.0𝑔
1 𝑚𝑜𝑙 𝑤𝑎𝑡𝑒𝑟
= 7.20 𝑔
% 𝑦𝑖𝑒𝑙𝑑 =
6.50 𝑔
7.20 𝑔
𝑥 100 = 90 %

3. Problem statement
For the reaction
2S(s) + 3O2(g) →2 SO3(s)
2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of
SO3 are collected. What is the percentage yield?

4. Problem statement
The Haber process can be used to produce ammonia, NH3, and it is based on the following
reaction.
N2(g)+3H2(g)→2NH3(g)
If one mole each of N2 and H2 are mixed and 0.50 moles of NH3 are produced, what is the
percent yield for the reaction?
calculate the theoretical yield
Given that 0.667 moles will produce 0.50 moles of NH3

5. Problem statement
What is the percent yield of the following reaction if 60 grams of CaCO3 is heated to
give 15 grams of
CaO+ CaCO3 → CaO + CO2
Theor. yield =
Now calculate the percent yield.

6. Problem statement
If the reaction of 77.0 grams of CaCN2 produces 27.1 grams of NH3, what is the percent
yield?
CaCN2+3H2O → CaCO3+2NH3
We know that CaCN2 and NH3's molar masses are 80 g/mol and 17 g/mol, respectively,
therefore

7. So, the percent yield, defined as the actual yield divided by the theoretical yield and
multiplied by 100%, is

8. Problem statement
What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3
grams of O2?
4C3H9N + 25O2 => 12CO2 + 18H2O + 4NO2
From C₃H₉N: considering CO2 is a product
From O₂:
The O₂ gives the smaller amount of CO₂, so the O₂ is the limiting reactant.

9. So, the percent yield, defined as the actual yield divided by the theoretical yield and
multiplied by 100%, is

10. Problem statement
At 273.15 K and 100kPa, 58.34 g of HCl reacts with 0.35 mol of MnO2 to produce 7.056
dm3 of chlorine gas. Calculate the theoretical yield of chlorine.
Which one of the two reactants acts as a limiting reactant
more moles of hydrochloric acid are present than need, therefore, Manganese dioxide
will act as a limiting reagent,

11. Theoretically, to produce one mole of chlorine gas for every mole of manganese dioxide
would reacts.
to calculate the percent yield of the reaction