Chemistry 112
First Hour Exam
Name:____________
Please show all work for partial credit
1. (10 points) Perform the following calculations and report your answers with the correct
number of significant figures
a. 23.7/0.0056 = 4,232.14... .0056 has 2 sig fig so round to 2 sig fig: 4200
b. 56.7× 0.1056 =5.98752... 56.7 has 3 sig fig so round to 3 sig fig: 5.99
c. 0.2355 -.3 =
.2355
-.3
-.0645 But digits past the tenths place are incorrect so:
-.1
d. 2.0x10-3 -.001 = .001 or 1x10-3
e (2300-576)/873.5 = 1.97367
2300
- 576
1724, but digits past the hundreds place are incorrect so:
1700 with 2 sig fig is the correct way to express this number
Final answer should therefore have 2 sig fig, and you should round UP to 2.0
2. (10 points) Perform the following unit conversions.
a. 536 mm to nm
1 nm
1x10 − 3 m
536 mm ×
×
= 536x10 6 nm
−9
1 mm
1x10 m
b. 15 ounces to grams
15 oz ×
1 lbs
1 kg
1000 g
×
×
= 425 g
16 oz 2.2046 lbs
1 kg
c. 15 lbs/in2 to g/cm2
15 lbs
1 kg
1000 g
1 in
1 in
×
×
×
×
= 1.055 g cm 2
in 2
2.2046 lbs
1 kg
2.54 cm 2.54 cm
2
3. (10 points) Name the following compounds
a. MgBr2 Magnesium bromide
b. Li2SO4 Lithium sulfate
c. HBr Hydrogen monobromide or hydrobromic acid
d. N2F4 Dinitrogen tetrafluoride
e. ClF 3 Chlorine trifluoride
4. (10 points) Give the correct chemical formulas for the following compounds
a. potassium oxide K2O
b. copper(IV) sulfate Cu(SO4)2
c. nitrous acid HNO2
d. sulfur difluoride SF2
e. chlorine monoxide ClO
5. (10 points) Give the % composition for each of he elements in the compound Mg(OH)2
1 Mg = 1x24.305
= 24.305
2 O = 2x15.999
=31.998
2 H = 2x1.007
= 2.014
Total=58.317
%Mg = 24.305/58.317 = 41.677 % Mg
% O = 31.998/58.317 =54.869 %O
%H = 2.014/58.317 = 3.454 %H
Total of % is 100.000, close enough for me
3
6. (10 points) A compound has a molecular weight of 407.68, and it contains 6.62% aluminum
and 93.38% Iodine. What is the molecular formula of the compound?
1 mole of compound will weigh 407.68 g so
1 mole of compound will contain
407.68g x .0662 = 26.99g Al
407.68g x .9338 = 380.69 g I
26.99 g Al = 26.99 g /(26.98 g Al/mol Al) = 1 mol Al
380.69 g I = 380.69/(126.9 g I/mol I) = 3 mol I
So molecular formula is AlI3
7. Balance the following chemical equation:
2NH3(g) + 5/2O2(g) 6 2NO(g) + 3H2O(g)
H
3
2
X3
adjust x2
N
2
1
x2
adjust
2
2
3
x5/2
adjust
Multiply entire equation by 2 to get rid of the fraction
4 NH3(g) + 2O2(g) 6 4NO(g) + 6H2O(g)
8. I am going to perform the following reaction in the lab:
C3H8(g) + 5O2 6 3CO2(g) + 4H2O(g)
a. If I have 5 g of C 3H8, how many grams of O2 will I need for this reaction?
5 g C 3 H8 ×
1 mole C3 H8
5 mole O2
32 g O2
×
×
= 1818
. g O2
44 g C3 H8
1 mole C3 H 8 1 mole O2
b. In I have 5 g of C 3H8 and an excess of O2, how many grams of CO2 will this reaction
produce?
Excess O2 insures that C 3H8 is limiting reagent so:
5 g C3 H8 ×
1 mole C3 H 8 3 mole CO2
44 g CO2
×
×
= 15 g CO2
44 g C 3 H 8
1 mole C3 H 8 1 mole CO2
4
9. Consider the reaction:
Mg(s) + I2(s) 6 MgI2(s)
If I have 2 g each of Mg and I2, which is the limiting reagent?
2 g Mg ×
1 mole Mg
1 mole MgI 2
×
= .0823 mole MgI 2
24.305 g Mg
1 mole Mg
2 g I2 ×
1 mole I 2 1 mole MgI 2
×
= .00788 mole MgI 2
2538
. g I2
1 mole I 2
I2 yields smaller amount of product so it is the limiting reagent
10. Aluminum burns in bromine in the following reaction:
2Al(s) + 3Br2(l) 6 2AlBr3(s)
If I start with 7.0 g of aluminum and an excess of Bromine, and I have a final yield of
50.3 g aluminum bromide, what was the percent yield for my reaction.
Excess Br2 so Al must be limiting reagent
7 g Al ×
1 mole Al
2 mole AlBr3 266.68 g AlBr3
×
×
= 69.188 g AlBr3
26.981 g Al
2 mole Al
1 mole AlBr3
% yield = 50.3g/69.188g x 100% = 72.7% yield