CHM1310-001
Fall 2000
Exam #1 (100 pts.)
Name
(Please print)________________________________
1.
(10)
Look at the periodic table in the front of the room. Write the names of the first 20 elements in order of increasing
atomic number.
Hydrogen,
Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon,
Sodium, Magnesium, Aluminum, Silicon, Phosphorous, Sulfur, Chlorine, Argon,
Potassium, Calcium
2.
(10)
Which prefix in the metric system would replace the following numbers given in
powers of 10?
10-3 milli
106 mega
10-6 micro
10-2 centi
103 kilo
3.
(8)
How many µg are in 0.0134 g?
(1.34
x 10-2 g) x 106 micrograms/g = 1.34 x 104
micrograms
4. (6) An elemental ion has 22 protons, 20
electrons, and 26 neutrons. What
isotope is it?
4826Fe
4726Fe
4822Ti
4822Ti2+ This one
4822Ti2-
5. (8) Use the periodic table to predict the product formed
when strontium (Sr) combines with chlorine.
SrCl2
6. The molecular formula of allicin, the compound responsible
for the characteristic smell of garlic, is C6H10OS2.
a)
(10)
How many moles of allicin are present in 5.00 mg of this substance?
Molar
mass = (6x12) + (10x1) + (1x16) + (2x32) = 162 g/mole
(5x
10-3 g) x (1 mole/152 g) = 3.09 x 10-5 moles
b)
(5) How many molecules of allicin are present in 5.00 mg of this
substance?
(3.29
x 10-5 moles) x (6.02 x 1023 molecules/mole) = 1.86 x 1019
molecules
c)
(5) How many sulfur atoms are present in 5.00 mg of this substance?
(1.86
x 1019 molecules) x (2 sulfur atoms/molecule) = 3.72 x 1019
sulfur atoms
7. (16) Calcium hydride reacts with water to form calcium
hydroxide and hydrogen gas. Given the
following unbalanced equation:
CaH2 + 2H2O --->
Ca(OH)2 + 2H2
How many grams of Ca(OH)2
are formed if 5.0 g of CaH2 are reacted?
(5.0
g CaH2) x (1 mole CaH2/42 g CaH2) = 0.12 moles
CaH2
(0.12
moles CaH2) x (1 mole Ca(OH)2/1 mole CaH2) =
0.12 moles Ca(OH)2
(0.12
moles Ca(OH)2) x (74 g Ca(OH)2/mole Ca(OH)2) =
8.9 g Ca(OH)2
8. One of the steps in the commercial process for converting ammonia
to nitric acid involves the conversion of NH3 to NO:
4NH3(g) + 5O2(g)
--> 4NO(g) + 6H2O(g)
If 2.50 g of NH3 reacts
with 2.85 g of O2
a) (16)
Which reactant is the limiting reactant?
(Show your work to justify your answer for credit.)
If
all the ammonia reacts how much NO is formed:
(2.50
g NH3) x (1 mole NH3/17 g NH3) = 0.147 moles
NH3
(0.147
moles NH3) x (4 moles NO/4 moles NH3) = 0.147 moles NO
(0.147
moles NO) x (30 g NO/mole NO) = 4.41 g NO
If
all the oxygen reacts how much NO is formed:
(2.85
g O2) x (1 mole O2/32 g O2) = 0.0891 moles O2
(0.0891
moles O2) x (4 moles NO/5 moles O2) = 0.0713 moles NO
(0.0713
moles NO) x ((30 g NO/mole NO) = 2.14 g NO
O2
is limiting reactant because it leads to less product. Therefore O2 runs out before ammonia
does.
b)
(6) Calculate the theoretical yield of NO in grams.
2.14
g NO from above calculation