Chemistry
Stoichiometry
Science and Mathematics
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Justification: The molar mass of HCl is 36.5 g/mol. Thus, the calculation to convert grams to moles is:
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Chemistry
Stoichiometry
Science and Mathematics
Education Research Group
Supported by UBC Teaching and Learning Enhancement Fund 2012-
D e p a r t m e n t o f C u r r i c u l u m a n d P e d a g o g y
a place of mind^ F A C U L T Y^ O F^ E D U C A T I O N
Stoichiometry
mL of HCl
L of HCl
mol HCl mol H
L of H 2
Solution
Answer: B
Justification: To balance the chemical equation, you need to determine how many of each part you have on either side of the equation and multiply by coefficients to make all parts equal on both sides.
The balanced chemical equation is thus:
Though nothing is written in front of Na 3 PO 4 , the coefficient of 1 is implied. The coefficient is not 0, as that would mean that no Na 3 PO 4 would be used.
H: 1 Cl: 1 Na: 3 PO 4 : 1 H: 3 Cl: 1 Na: 1 PO 4 : 1 x3 x
A. 0.37 mol
B. 1.6 x 10^2 mol
C. 0.12 mol
D. 3.0 mol
E. None of the above
Stoichiometry II
If you have 4.5 g of HCl, how many moles of HCl would you have?
Stoichiometry III
Knowing that you have 0.12 mol of HCl, what conversion factor would you use to determine how many moles of H 3 PO 4 you would produce?
Solution
Answer: E
Justification: To determine the amount of product produced from a reactant, you need to use the mole ratio between the reactant and product as a conversion factor.
A and D are incorrect because they are the molar mass of H 3 PO 4. To convert from moles of a reactant to moles of a product, you need the mole ratio not the molar mass.
Continued on next slide…
A. 3.0 mol
B. 0.04 mol
C. 1.0 mol
D. 0.12 mol
E. None of the above
Stoichiometry IV
Knowing that you have 0.12 mol of HCl, how many moles of NaCl would you produce?
Solution
Answer: D
Justification: The mole ratio between HCl and NaCl is a 3: ratio (or 1:1). Thus the amount of moles of HCl that you start with would be the same as the amount of moles of NaCl that would be produced.
You would get B if you used the same 3:1 mole ratio from the last question.
A and C are incorrect because the values in the mole ratios do not tell you how many moles will be produced. Rather they tell you how much product relative to another substance will be produced.
Solution
Answer: E
Justification: To do the calculation in one step you would multiply all of the individual conversion factors and mole ratios together one after the other.
All of these answers will get the correct answer, however the best way of writing out the steps was not an option. The best practice method is shown below:
Notice how each fraction is a different conversion factor.
Continued on next slide…
Solution
Answer: E
Justification: It is important to use the proper method when writing out stoichiometry calculations so that it becomes easier to see what conversion factors were used, how the units cancel out and to identify which compound each conversion factor refers to (since the compounds are explicitly written in).
Notice that in the other answers, one or both of these objectives is not achieved.
Solution
Answer: B
Justification: D and E are incorrect because to determine the amount of product produced from the amount of a reactant. In questions like this, you always need to go through the mole ratio. After that you can use the molar mass of the product to determine the mass in grams produced.
Instead of using a 4:2 ratio, you could also simplify it to a 2:1 ratio. A is incorrect however, because the ratio is inverted and the unit would not cancel.
To solve this question the sequence is:
mol H 2 mol HCl mass of HCl
Stoichiometry VII
What conversion factors (in the correct order that you would multiply them) would you use in order to determine how many liters of H 2 gas you would need to react to produce 25 g of HCl (at STP)?
SiCl 4 + 2H 2 Si + 4HCl
A. Mass of HCl / mole ratio between HCl and H 2 / molar mass of H 2
B. Molar mass of HCl / molar mass of H 2 / molar volume of H 2
C. Mass of HCl / mole ratio between HCl and H 2 / molar volume of H 2
D. Molar mass of HCl / mole ratio between HCl and H 2 / molar volume of H 2
E. None of the above
Stoichiometry VIII
Rust (Fe 2 O 3 ) forms on iron according to the above unbalanced equation. How much iron would have had to be reacted to form 10.0 g of rust?
Fe + O 2 Fe 2 O 3
A. 3.50 g
B. 14.3 g C. 10.0 g
D. 7.0 g E. None of the above
Solution
Answer: D
Justification: The balanced chemical equation is:
To determine the mass of Fe, the mass of rust had to be converted as shown below:
g of rust mol of rust mol of Fe g of Fe
Continued on next slide…