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1.
Question 1
Ammonia (NH
3
), a gas used in the production of fertilizer, is one of the top 10
most produced chemicals in the world. In 2014, 1.76
10
11
kg of ammonia was
produced worldwide. How many moles of ammonia are in 1.76
10
11
kg of
ammonia? Show your work.
Explanation: To determine the number of moles of ammonia in 1.76 x 10^11 kg
of ammonia, you need to use the concept of molar mass and the formula Moles =
Mass / Molar mass. The molar mass of ammonia (NH3) is the sum of the atomic
masses of its constituent elements: nitrogen (N) with a molar mass of
approximately 14.01 g/mol and hydrogen (H) with a molar mass of approximately
1.01 g/mol. So, the molar mass of ammonia is around 17.02 g/mol. You'll need to
convert the mass from kilograms to grams, so 1.76 x 10^11 kg is equivalent to
1.76 x 10^14 grams. Then, you can use the formula to find the number of moles:
Moles of ammonia = 1.76 x 10^14 g / 17.02 g/mol = 1.03 x 10^13 moles.
2.
Question 2
A common pain-relieving medication contains acetaminophen and codeine. If a
1.5 g tablet contains 0.30 g acetaminophen (C
8
H
9
NO
2
, 151.16 g/mol) and 0.044 g
codeine (C
18
H
21
NO
3
, 299.36 g/mol), calculate the number of moles of codeine in
the tablet.
Explanation: You would use the mole concept. You find the molar mass of each
compound, then use the formula Moles = Mass / Molar mass to calculate the moles of
acetaminophen and codeine separately. This approach allows you to quantify the
number of moles of codeine in the tablet.
1.
1.5 × 10
-4
mol
3.
Question 3
Hydrogen peroxide decomposes to produce water and oxygen. Which relationship
regarding the quantities of reactants and products associated with this reaction is
correct?
Explanation: This equation indicates that for every 2 moles of hydrogen peroxide that
decompose, 2 moles of water and 1 mole of oxygen are produced. So, the correct
relationship is that two moles of water and one mole of oxygen are produced for every
two moles of hydrogen peroxide that react. This is important for stoichiometry and
determining the quantities of reactants and products in chemical reactions, ensuring
that the reaction obeys the law of conservation of mass.
Answer: The decomposition of two moles of hydrogen peroxide results in one mole of
oxygen gas.
4.
Question 4
Airbags in cars inflate when an electrical spark activates sodium azide (NaN
3
) so
it decomposes to sodium metal and nitrogen gas. In the balanced reaction
equation, how many moles of N
2
gas are formed for each mole of NaN
3
?
Explanation: In this balanced equation, you can see that for every 2 moles of
sodium azide (NaN3) that decompose, 3 moles of nitrogen gas (N2) are
produced. Therefore, the ratio is 3 moles of N2 gas for each mole of NaN3. This
balanced equation shows the stoichiometric relationship between the reactant
(NaN3) and the product (N2 gas) in the chemical reaction. Understanding this
ratio is essential for determining the amount of product formed in chemical
reactions and ensuring the conservation of mass.
Answer: 1.5
5.
Question 5
Copper(II) sulfate (CuSO
4
) can be used to control algae growth. Aqueous CuSO
4
solutions that come into contact with the surface of galvanized (zinc-plated) steel
pails undergo the following balanced reaction:
How many grams of zinc would react with 454 g (1 lb) of CuSO
4
(159.61 g/mol)?
Explanation: First, find the molar mass of CuSO4 (copper(II) sulfate), which is 159.61
g/mol. Then, use the balanced chemical equation to establish the stoichiometric ratio
between CuSO4 and Zn. Since the equation shows a 1:1 ratio between CuSO4 and Zn,
you can directly convert the given 454 grams (1 lb) of CuSO4 into moles, which is
approximately 2.84 moles. Since the ratio is 1:1, you know that 2.84 moles of CuSO4 will
also react with 2.84 moles of Zn. Finally, calculate the mass of zinc (Zn) by multiplying
the number of moles by its molar mass, which is 65.38 g/mol. This gives you a final
result of 185.63 grams of zinc required to react with 454 grams of CuSO4.
Answer: 186 g
6.
Question 6
Potassium superoxide (KO
2
, 71.10 g/mol) can be used to generate oxygen gas. If
the theoretical yield of O
2
(32.00 g/mol) must be 12.0 g, how many grams of KO
2
should be used? Assume the reaction occurs in the presence of excess H
2
O (18.02
g/mol).
Explanation:
To determine the grams of potassium superoxide (KO2) required to
produce a theoretical yield of 12.0 grams of oxygen gas (O2), we need to use
stoichiometry. First, we balance the chemical equation, which tells us that 4 moles of
KO2 produce 3 moles of O2. We convert the given 12.0 grams of O2 into moles using the
molar mass of O2. This gives us 0.375 moles of O2. We then use the mole ratio to find
the moles of KO2 needed, which is 0.5 moles. Finally, we convert these moles back into
grams using the molar mass of KO2 to get 35.55 grams of KO2 needed to achieve the
desired theoretical yield.
Answer: 35.6 g
7.
Question 7
Ammonia (NH
3
) can be produced by reaction of hydrogen gas (H
2
) with nitrogen
gas (N
2
). Write out the steps followed to determine the moles of NH
3
produced
when 56 grams of N
2
is reacted with 43 grams of H
2
.
Explanation and answer: First, you have to write a balanced equation: 3H
2
+ N
2
produces 2NH
3
. Then in order to determine the number of moles of NH
3
produced
from each reactant to see which reactant produces less you have to:
56 grams N
2
x (1 mol N
2
/28g N
2
) x 2 mol NH
3
/1 mol N
2
= 4 mol NH
3
.
43 grams H
2
x (1 mol H
2
/2 g H
2
) x (2 mol NH
3
/3 mol H
2
) = 14.3 mol NH
3
.
Recognize that N
2
is the limiting reactant and the number of moles of NH
3
produced is 4 moles.
8.
Question 8
To make your favorite omelet, you use 3 eggs, 1/2 cup cheese, 1/3 cup
mushrooms, and 1 Tablespoon of butter.
Which ingredient will limit the number of complete omelets that you can make?
Explanation and answer: The limiting ingredient will be what runs out first. You
need exactly those measurements to make 1 omelet.
9.
Question 9
H
2
O decomposes to form O
2
and H
2
. If the % yield for a given reaction is 85% and
the amount of O
2
obtained is 5.82grams, what is the theoretical yield in moles of
O
2
?
Explanation: To determine the theoretical yield of O2 in moles, first, acknowledge
that the % yield is 85%, which means the actual yield is 85% of the theoretical
yield. Given the actual yield of 5.82 grams of O2, calculate the theoretical yield
by dividing the actual yield by the % yield expressed as a decimal (0.85). This
gives a theoretical yield of approximately 6.85 grams. Next, convert this
theoretical yield from grams to moles by dividing by the molar mass of O2 (32.00
g/mol).
Answer: % yield = (actual yield/theoretical yield) x 100
85 = (5.82grams of O
2
/ theoretical yield of O
2
in grams) x 100
Theorietical Yield = (5.82/85) x 100 = 6.85g O
2
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