1.
What is the weight of a mole of carbon atoms if the gram molecular volume is defined as 33.6
liters?
A) 8.00 g
B) 12.0 g
C) 18.0 g
D) 24.0 g
E) 36.0 g
Explanation of the answer: In the problem # 1 above we are given 33.6 liters of carbon and asked to
find the weight (mass) in grams. To answer this problem correctly, we will convert the given 33.6 L to
moles by dividing it with 22.4 L because I mole of any gas at STP contains 22. 4 L. Then we will convert
the calculated mole(s) into gram by multiplying it with 12, the atomic mass of carbon. By using the
following dimensional analysis calculation
𝟏 𝒎𝒐𝒍𝒆
33.6 L x
𝟐𝟐.𝟒 𝑳
𝑿
𝟏𝟐 𝒈 𝒐𝒇 𝑪
𝟏 𝒎𝒐𝒍𝒆
= 𝟏𝟖. 𝟎 g we can say that 18.0 g of carbon occupy a volume of 33.6 liters or vice versa.
The mass of one mole of a gas with a density at STP of 2.9 g•L–1 is close to
2.
A) 2.9 g
B) 24 g
C) 44 g
D) 64 g
E) 76 g
Explanation of the answer: In the problem # 2 above we are asked to find the mass in grams for the gas with
𝑔
density of 2.9 g.L-1, which can also be written as 2.9 𝐿 . The Density formula is given in Table T which is D=
which can be rearranged for the calculation of mass as follow:
𝑚
𝑉
,
M= D*V
We know that a mole of a gas at STP occupies 22.4 liters of volume. Therefore, the setup to solve the given
problem will be
𝑔
M= 2.9 𝐿 X
22.4 𝐿
1 𝑚𝑜𝑙𝑒
= 64.96
𝑔
1 𝑚𝑜𝑙𝑒
or
Mass of 1 mole = 64.96 g
3. Seven grams of a gas occupies 11.2 liters at one atmosphere pressure and 546 K. The gas could be
B) O2 C) CO
D) NH3 E) N2
A) Ar
4. A 1.00 L sample of a gas, at standard conditions, has a mass of 1.96 g. The gas is most likely to be
a.
H2
B) N2
C) O2
D) CO2
E) Ar
Explanation of the answer: To solve the problem # 4 first we will find the formula mass of each molecule.
H2=1x2= 2 g/mole N2=14x2=28g/mole
O2=16x2=32g/mole CO2=(1x12)+(16x2)=44g/mole Ar= 1x36=36g/mole
Then we convert the given volume of 1.00 L into moles,
1 𝑚𝑜𝑙𝑒
1.00 L 𝑋 22..4 𝐿! =0.0446 moles
We learned that the given mass in grams can be converted into mole by dividing it with GFM by using
𝑔𝑖𝑣𝑒𝑛 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
the following equation:
moles= 𝐺𝐹𝑀 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑀𝑀 . Rearranged equation for the calculation of
GFM or MM will be
𝐺𝐹𝑀 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑀𝑀 =
𝑔𝑖𝑣𝑒𝑛 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
𝑚𝑜𝑙𝑒𝑠
.
Now let us plugin the given above mass (1.96 g) and the calculated moles (0.0446) in the rearranged
1.96 𝑔
𝐺𝐹𝑀 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑀𝑀 = 0.0446 𝑚𝑜𝑙𝑒𝑠 = 44g/mole
equation to calculate GFM or MM.
44 g/mole is the GFM or MM of CO2. Therefore, the answer is choice D.
Hydrogen gas, H2(g), reduces hot iron oxide (MM = 232), to iron metal.
5.
Fe3O4(s) + 4 H2 (g)  3 Fe (s) + 4 H2O(g)
How many moles of water are formed by passing dry H2 over 464 g of Fe3O4 until reduction is complete?
A) 16
B) 2.0
C) 8.0
D) 4.0
E) 5.0
Explanation of the answer: To solve, mass-mole stoichiometry problem, the problem # 5, first we will
convert the given grams (464g) of Fe3O4 to moles by dividing it with the gram formula mass (GFM) of 232
g/mole and then we will use the mole ratio given in the equation for water and Fe3O4 to calculate the moles
of water formed
𝟏𝒎𝒐𝒍𝒆 𝒐𝒇𝐅𝐞𝟑𝐎𝟒
𝟒𝟔𝟒 𝒈 𝒐𝒇 Fe3O4𝑿 𝟐𝟑𝟐 𝒈 𝒐𝒇 𝐅𝐞𝟑𝐎𝟒 𝑿
𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒔 𝟒 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓
𝟏 𝒎𝒐𝒍𝒆 𝒐𝒇 𝐅𝐞𝟑𝐎𝟒 𝐢𝐧 𝐭𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
= 8.0 moles of water
Answer
6.
When 13.0 grams of acetylene, C2H2, is reacted, what mass of water, H2O, is produced?
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
A) 9.00 g
B) 12.0 g
C) 13.0 g
D) 18.0 g
E) 26.0 g
Explanation of the answer: To solve, mass-mass stoichiometry problem, the problem # 5, first we will find
the gram formula mass (GFM) for C2H2 and H2O.
C2H2 = (12x2)+(1x2)=24+2=26g/mole
H2O =(1x2)+(16x1)=2+16=18g/mole
Now we will convert the 13.0 g of acetylene, C2H2, to moles and then use it along with the mole ratios
given in the equation to solve the problem.
13.0 𝑔 𝑜𝑓 C2H2𝑋
=
1𝑚𝑜𝑙𝑒 𝑜𝑓C2H2 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑠 2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
18 𝑔
𝑋
𝑋
26 𝑔 𝑜𝑓C2H2
2 𝑚𝑜𝑙𝑒 𝑜𝑓 C2H2 in the equation
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠
9.00 g
Answer