Problem Set Stoichiometry Answers
1) Draw a picture showing the following reaction:
12 molecules of hydrogen gas as a reactant
3 molecules of nitrogen gas as a reactant
The maximum molecules of ammonia (NH3) gas that can be produced
Use blue for hydrogen and red for nitrogen.
Give the balanced synthesis reaction.
Remember your diatomic elements.
+
3H2 (g) + N2 (g)  2NH3 (g)
There are 3 molecules of hydrogen that do not react – they are in excess.
2) Define “Reactant in Excess” and “Limiting Reactant.” Why are these two terms
important in industrial production of compounds?
The reactant is excess is the substance that remains after the limiting reactant runs
out. The limiting reactant is the reactant that runs out before any other in a
chemical reaction – it limits the amount of product made. The limiting reactant in
most industrial reactions are typically the most expensive compounds.
3) Explain why “Theoretical Yield” is usually more than the “Actual Yield” for most
reactions.
The theoretical yield is typically more than actual yield because of human errors,
laboratory conditions are not always ideal, the concentrations of products slows
reactions down by getting in the way… others.
4) Determine the Percentage Yield if a reaction has the theoretical yield of 3.45
grams of product and the actual yield of 3.25 grams of product.
3.25 g / 3.45 g = 94.2 % yield
5) Determine the actual yield of a reaction that has a 77 % yield and a theoretical
yield of 2.35 grams.
Actual Yield / 2.35 g = 77%
Actual = (.77)(2.35g) 1.81 g
6) Determine the theoretical yield of a reaction if it has a 85% yield and an actual
yield of 3.22 grams.
3.22 g / Theoretical Yield = 85% Theoretical = 3.22 / 0.85 3.79 g
Use this reaction for the next four questions:
Cu (s) + 4 HNO3 (aq)  Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)
7) 3.0 moles of copper react completely in a reaction.
a. How many moles of nitric acid are needed?
3.0 mol Cu x (4 mol HNO3 / 1 mol Cu) = 12.0 mol HNO3
b. How many moles of copper (II) nitrate are produced?
3.0 mol Cu x (1 mol Cu(NO3)2 / 1 mol Cu) = 3.0 mol Cu(NO3)2
c. How many moles of water are produced?
3.0 mol Cu x (2 mol H2O / 1 mol Cu) = 6.0 mol H2O
d. How many moles of nitrogen dioxide are produced?
3.0 mol Cu x (2 mol NO2 / 1 mol Cu) = 6.0 mol NO2
e. How many liters of NO2 are produced if the lab was performed at STP?
OMIT THIS QUESTION - 6.0 mol NO2 x 22.4 L/ 1mol = 134.4 L
8) If 1.2 moles of copper react with 4.0 moles of HNO3, what is the limiting
reactant?
HAVE
NEED
1.2 mol Cu x (4mol HNO3 / 1 mol Cu) = 4.8 mol HNO3
4.0 mol HNO3 x (1 mol Cu / 4 mol HNO3) = 1 mol Cu
You don’t have enough HNO3 so HNO3 is the limiting reactant.
9) If 1.2 moles of copper are placed into 4.9 moles of HNO3, how much of the
reactant in excess remains at the end of the reaction?
HAVE
NEED
1.2 mol Cu x (4mol HNO3 / 1 mol Cu) = 4.8 mol HNO3
4.9 mol HNO3 x (1 mol Cu / 4 mol HNO3) = 1.22 mol Cu
Cu limits the reaction and there is 0.10 mol of HNO3 left over. (4.9 – 4.8)
10) How many grams of water can 12.5 grams of anhydrous chromium (III) sulfate
salt absorb if the formula of the hydrated salt is Cr2(SO4)3.18H2O?
Cr2(SO4)3.18H2O (s)  Cr2(SO4)3 (s) + 18H2O (g)
12.5 g Cr2(SO4)3 x (1 mol /392 g ) = 0.0319 mol Cr2(SO4)3
0.0319 mol Cr2(SO4)3 x (1 mol /392 g ) = 0.0319 mol Cr2(SO4) 3.18H2O
0.0319 mol Cr2(SO4) 3.18H2O x ( 716 g/1 mol) = 22.8 g Cr2(SO4) 3.18H2O
11) What mass of magnesium oxide should be produced from 3.55 grams of
magnesium and an excess of oxygen gas?
2Mg (s) + O2 (g)  2MgO (s)
3.55 g Mg x (1 mol /24.3 g Mg) = 0.146 mol Mg
0.146 mol Mg x (2 mol MgO/ 2 mol Mg) = 0.146 mol MgO
0.146 mol MgO (40.3 g MgO/ 1 mol MgO) = 5.88 g MgO
12) How many grams of copper will be produced from reacting 3.44 grams of iron
with 3.55 grams of Cu(NO3)2 that have been dissolved in water? (you need to
determine the limiting reactant)
3 Cu(NO3)2 (aq) + 2 Fe (s)  2 Fe(NO3)3 (aq) + 3 Cu (s)
3.44 g Fe x (1 mol Fe/ 55.85 g Fe) = 0.0615 mol Fe (HAVE)
3.55 g Cu(NO3)2 ( 1 mol/187.55 g) = 0.0189 mol Cu(NO3)2 (HAVE)
0.0615 mol Fe x (3 mol Cu(NO3)2 / 2 mol Fe) = 0.09225 mol Cu(NO3)2 NEEDED
0.0189 mol Cu(NO3)2 x (2 mol Fe/3 mol Cu(NO3)2 ) = 0.0126 mol Fe NEEDED
You have enough Fe, you don’t have enough Cu(NO3)2.
0.0189 mol Cu(NO3)2 x (3 mol Cu/3 mol Cu(NO3)2 ) = 0.0189 mol Cu produced
0.0189 mol Cu x (63.5 g Cu/ 1 mol Cu) =1.20 g Cu can be made.
13) What mass of CO2 gas will be produced from the combustion of 50.00 grams of
octane, C8H18?
2 C8H18 (g) + 25 O2 (g)  16 CO2 (g) + 18 H2O (g)
50.00 g C8H18 x (1 mol/114 g) = 0.439 mol C8H18
0.439 mol C8H18 x ( 16
mol CO2 /
2 mol C8H18 )
=0.055 mol CO2
0.055 mol CO2 (44.0 g/1 mol) = 2.41 g CO2
Final Exam Format and Expectations
For each of the labs you complete for the exam please complete the following:
1) Follow the directions given.
2) Record your observations in your lab book in an organized fashion. (10 pts each)
3) Record your data in your lab book in an organized fashion. (10 points each lab)
Once you have completed the first three sections, complete the data analysis portion
alone at your lab bench. (30 points each)
There is a 4th section to be completed at the lab bench on Thursday alone at the lab bench
once the first three sections are completed.
You will need to be able to:
Name ionic compounds.
Write the formulas of ionic compounds.
Determine percent composition based upon empirical formulas
Determine percent composition based upon data collected.
Determine molar ratios of water in a hydrate.
Write balanced chemical equations for reactions
Predict the products of single and double replacement reactions.
Convert from grams to moles of substances.
Convert from moles to grams of substances.
Convert from moles of one substance to moles of another substance.
By Thursday 3:00 exams are finished, unless you missed a class day.
By Friday 3:00 exams and all missing work is turned in and completed.
Turn in your text book this week.
Turn in your lab book this week.
Good Luck!