Download 1617 Ch3 Practice Test Key Student

AP Chemistry Take-Home Test Key: Stoichiometry
1. Determine each of the following for a 30.0 g sample of Al(OH) 3.
A) the amount of moles for the sample.
Use the molar mass of Al(OH)3 (78.01 g/mol) to convert grams to moles
Answer: 30.0/78.01 = 0.385 mol Al(OH)3
B) the amount of formula units (molecules) in the sample.
“Formula units” are the things, and “counting things” is related to Avogadro’s number: 1 mole = 6.02  1023 things counted.
Answer: 0.385(6.02  1023) = 2.32  1023 formula units of Al(OH)3
C) the amount of hydroxide ions in the sample.
The total OH- ions in the sample depends on the ratio of Al(OH)3 : OH-, which is 1 Al(OH)3 : 3 OH- and on how many units of
Al(OH)3 are in the sample (the answer to part B).
Answer: 3(2.32  1023) = 6.96  1023 OH1- ions
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2. Lysine, an essential amino acid in the human body (amino acids are the building blocks of proteins), contains only the elements
C, H, O, and N. Different experiments, each using 2.175 g of lysine, were conducted to determine its formula.
A) If complete combustion of the 2.175 g sample produced 1.89 g H 2O, determine the mass of H in the
sample.
To determine g H in H2O, you need a mass ratio of H : H2O. With two H’s in H2O, the mass ratio is 2.02 g H : 18.02 g H2O.
 2.02 g H 
  0.212 g H
 18.02 g H 2 O 
Answer: 1.89 g H 2 O 
B) When the 2.175 g sample reacted completely in a different experiment, 0.507 g NH3 was produced. Determine the mass of N in
the sample.
Similar to part A, except that you are using NH3 and N. With one N in NH3, the mass ratio is 14.01 g N : 17.04 g NH3.
 14.01 g N 
Answer: 0.507 g NH 3 
  0.417 g N
 17.04 g NH 3 
C) A separate test showed that lysine is 49.4% C. Determine the mass of C and O in the 2.175 g sample of lysine.
*Since you know the mass percent of C, determine the g C in the given sample by multiplying the given sample by the percent.
*Since the samples used in parts A, B, and C have the same mass, it is like you’re really using only one sample: one that weighs
2.175 g. Thus, the g O is the difference between the mass sample and the mass of each element previously determined.
Answer: (0.494)(2.175) = 1.07 g C
2.175 – (0.212 + 0.417 + 1.07) = 0.476 g O
D) Determine the empirical formula for lysine.
Each subscript in the formula is determined by the number of moles. Convert the masses to moles and ensure that each subscript
is also a whole number by dividing and if necessary, multiplying.
Answer: 1.07/12.01 = 0.0891 mol C
0.212/1.01 = 0.210 mol H
0.417/14.01 = 0.0298 mol N
0.476/16.00 0.0298 mol O
C0.0891H0.210N0.0298O0.0298 = C3H7NO
E) Lysine’s molar mass is about 150 g. Determine the molecular formula of lysine.
The mass of the empirical formula must be compared to the mass of the true formula. If you have done everything correctly, the
molar mass should be a whole number multiple (1, 2, 3, etc) of the empirical formula. Multiply the empirical formula by this
multiple (if necessary) to get the molecular formula.
Answer: mass of C3H7NO  73 g
150/73  2
C6H14N2O2
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3. Nitrogen gas, N2, can be prepared by reacting copper (II) oxide with ammonia gas, NH 3. Along with production of N2, the
reaction also produces solid copper, Cu, and gaseous water.
A) Write a complete balanced equation for the reaction described.
The correct formula for copper (II) oxide must be written first (from Cu 2+ and O2-) and all compounds must be placed on the
correct side of the chemical equation. From the description, N2, Cu, and H2O are products while CuO and NH3 are reactants.
Answer: 3CuO + 2NH3  N2 + 3Cu + 3H2O
B) Determine the moles of NH3 needed to produce 20.0 g N2.
Use molar mass to determine moles of N2 produced. Then use the ratio in the balanced equation to determine moles of NH3.
Answer: 20.0/28.02 = 0.714 mol N2 produced
Ratio of 1 N2 : 2 NH3, therefore 2(0.714) = 1.43 mol NH3 needed
C) Assume 18.1 g of NH3 reacts with 90.4 g of copper (II) oxide.
i) Determine the moles of each reactant.
Convert each of the masses using the molar mass of each compound: 17.04 g/mol for NH 3 and 79.55 g/mol for CuO.
Answer: 18.1/17.04 = 1.06 mol NH3
90.4/79.55 = 1.14 mol CuO
ii) Determine the limiting reactant. Clearly justify your answer with calculations or in words.
One way to determine the limiting reactant (as explained in class) is to choose one given reactant and use it to determine how
much of the other reactant is needed. Then compare this needed amount to the given amount of the “other”. You should arrive at
the same conclusion regardless of which reactant you choose to start with. Your textbook explains an alternative method of finding
the limiting reactant.
 3 mol CuO 
Answer: 1.06 mol NH3 
  1.59 mol CuO needed
 2 mol NH3 
1.59 mol CuO is needed, but only 1.14 mol is given. Therefore, CuO is limiting.
iii) Determine the mass of N2 produced in the reaction.
The given limiting reactant determines everything that happens in the reaction! Use the given amount of limiting reactant, along
with the balanced equation and molar mass, to answer the question.
 1 mol N 2   28.02 g N 2   10.6 g N
Answer: 1.14 mol CuO 

2

 3 mol CuO   1 mol N 2 
iv) Determine the moles of excess reactant remaining once the reaction is complete.
The given limiting reactant determines everything that happens in the reaction! Use the amount of limiting reactant to determine
the amount of the excess reactant needed, and then subtract this amount from the given.
 2 mol NH3   0.760 mol NH needed
Answer: 1.14 mol CuO 

3
 3 mol CuO 
1.06 – 0.760 = 0.30 mol NH3 remaining
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4. Hydrogen and oxygen react according to the reaction below.
2H2 + O2  2 H2O
Using particulate diagrams, represent the initial and final amounts of substances when 8 molecules of H 2 react with 5 molecules of
O2.
Represent each initial amount appropriately. Then, using the balanced equation, determine the limiting reactant, the amount of H2O
produced, and the amount of excess remaining at the end.
Answer: