CKftrA. II Exam Review: Calculations PK<3r#l
Date: HAV Room: S12 Time: "75 /Yliaa'feS Marks: H oP % of grade:^Q%
You wii! be given a periodic table (as used for tests), a list of polyatomic ions, solubility rules,
and a list of constants: 6.02x1023, 22,4 L, 245 L, 273 K, 101.3 kPa, 8.31 (kPa-L)/fmol K)
Basically, any information that you did not have to memorize for a test, you do not have to memorize for the exam.
This sheet is not a complete review of 6tr« M chemistry. You will still need to review ail of your notes (especially
focus on unit review sheets). This review is meant to help you sort through the different types of calculations that we
have done. Try to solve these questions first without your notes (after all, you won't have your notes during the exam).
1. How many electrons, protons, and neutrons are in i9F? Ar-40?
2. What chemical formula results when the following elements are combined: a) Al + S, b) Mg + 0?
4. Calculate the percentage composition (by mass) of H3PO4.
5. 4.89 -j- 6.77 + 1.2782 x 2.78 = (give your answer with the correct number of significant digits).
6. Boron has two isotopes: 10B (19.9%) and 11B (80.1%). Calculate Boron's average atomic mass.
7. How many atoms are in 7.3 moles of C2H4?
8. How much does 3.8 mol CUSO4 weigh in grams?
9. A compound contains C (63.2% by mass), H (8.8%), and 0 (28.0%). Calculate its simplest formula.
10. A compound has the simplest formula CH2O and a molar mass of 180 g/mol. Give its molecular formula.
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12. Butane burns according to the equation: 2C4H1o + 1302 8CO2 + IOH2O. If a 10 mL test tube is filled
1/10 with butane and 9/10 with O2, how much 62 will remain at the end of the reaction?
13. 2KNO3 -» 2KNO2 + O2. What mass of O2 can be produced from 592 grams of KNO3?
14. 3Fe + 2O2 Fe304. What mass of Fe304 can be produced if 100 grams of O2 reacts with 4.91 mol Fe?
15. CuS04(aq) + BaCl2(aq) CuCl2(aq) + BaS04(s). What is the percentage yield of BaS04, if adding
18.0 g of BaCh to excess CUSO4 produces 13.2 grams of BaSC^?
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18. What is the molar concentration of a solution that holds 16 g of NaOH in 2.00 L of solution?
19. What volume of 12.0 M HCI is needed to make 3.00 L of a 0.175 M solution?
20. 3.5 L of 1.3 M KCI is added to 2.0 L of 0.75 M KCI. What is the concentration of the resulting solution?
22. What mass of precipitate forms when 0.35 L of 0.175 M CaCh is added to excess NasPC^aq)?
24. In a titration, 4.42 mL of 0.600 M H2SO4 neutralized 5.00 mL of NaOH. What was the concentration of NaOH?
25. a) 22 K = 0C b) 7560C = K
26. A aerosol can at 250C has an internal gas pressure of 150 kPa. What will the pressure be at 200oC?
27. A weather balloon holding 22 L of gas is released at a pressure of 103 kPa & a temperature of 150C. What
is the temperature at 10 km, where the pressure is 15 kPa and the balloon occupies a volume of 108 L?
28. A gas is collected in a long glass tube over water. After equalizing the water level inside and outside the
tube, the volume was measured at 35.7 mL. If the atmospheric pressure is 97 kPa and the water
temperature is 20.0oC (vapour pressure = 2.34 kPa), calculate the volume of dry gas at STP.
29. A 2.62 g sample of a gas occupies 1.46 L at 220C and 110 kPa. What is the molar mass of the gas?
30. 2H2 + 02-» 2H2O. What mass of water is produced when 2.4 L of O2, at RjP, is reacted with excess H2?
31. Write a balanced equation for the complete combustion of a 27-carbon hUdhTo Gxft^: ^5^
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1. 19F: 9e- 9p+1 10n0 Ar-40:18e-, 18p+) 22n0
2. a) AI2S3 b) MgO
3. NaF = 4.0 - 0.9 = 3.1: ionic, London.
H2O = 3.5-2.1 = 1.4: dipole-dipole
(specifically H-bonding), London.
O2 = 3.5 - 3.5 = 0: London
4. H: 3.03/98x100 = 3.1% H
P: 30.97/98x100 =31.6%P
8. # g CUSO4 = 3.8 mol x 159.1 g/mol = 606 g
9. 63.2 g C x 1 mol/12.01 g = 5.26 mol (3.01)
8.8 g H x 1 mol/1.01 g = 8.71 mol (4.98)
28.0 g 0x1 mol/16.0g= 1.75 mol (1)
Simplest formula = CsHgO
10.CH2O; 30 g/mol, 180 g/moi + 30 g/mol = 6
Molecular formula: C6H1206
11.21083Bi^42He + 20681TI
0:64/98x100 = 65.3% O
7534Se ^ ^e + ^Br
12.#mL02 = 1 mLC4H10 x 13mLO2/2mLC4H10
5. 4.89 + 6.77 + 1.2782 x 2.78
= 0.7223 + 3.5534 = 4.2157 =4.28
= 6.5 mL used, 2.5 mL remain
6. (0.199)(10) + (0.801)(11) = 10.801
13. #g02= = 93.7g02
7.7,:
, r_ 1 mmm* r 32.00 gOa
1
= 2.64X1025 atoms
3 101.110^3 2O»H^I03 1jBeTO2
14. #gFe304=
1 1 231.55 g Fe304
2 mot 02 1 m
382 g Fe304
.911«MU« x x2f55!F;£f4s
'4
-3U4
5. # g BaS04=
1 meHaQla met-BaSQ-4 x 233.39 g BaSO,
208.23 gSa^-1 1
17. ppm = mg/kg, 4.0 ppm = x mg / 50000 kg
x = 200 000 mg = 200 g = 0.20 kg
18. # mol = 16g x 1 mol / 40g = 0.40 mol NaOH
[NaOHJ = 0.40 mol / 2.00 L = 0.20 M
lO.M^ = M2V2
(12.0 M)^) = (0.175 M)(3.00 L)
V, = 0.04375 L = 43.8 mL
20.3.5Lx1.3mol/L + 2.0Lx0.75mol/L = 6.05mol
351 + 201 = ^ !
mol/L = 6.05 mol/5.5 L = 1.1 mol/L
21 .g/100 mL H20: at 520C = 90, at 20oC = 32.
% yield = actual-5-theoretical x 100%
= 13.2g- 20.175 gx 100% = 65.
16. 15 g 42 g x 100% = 35.7%
22. First, we need a balanced equation:
2Na3P04(aq) + 3CaCI2(aq)-+ Ca3(P04)2(s)+ 6NaCI(aq)
«¦ y ^4/2
oss-t 9A1 JIOgCa^PQ^
1-t 3 mol Cc€I2 1 molCa^PO^
90 - 32 = 58 g/100 mL H20 precipitates
58 g/100 mL x 70 mL = 40.6 g precipitate
26. = 150 kPa, T1 = 298 K
P2 = ? kPa, T2 = 473 K, V^
P2 = 238 kPa
pH = -log[H+] = -Iog[1.85x10-12| =11.7
24. MaxVaX#H = MbXVbX#OH
(0.600 mol/L)(4.42 mL)(2) = (MB)(5.00 mL)(1)
Mb = 1.06 mol/L
25. K = "C - 273, or 0C = K + 273
T2 = 206 K = -670C
30. #gH20=
Tc
28. P1 = 97 kPa - 2.34 kPa = 94.66 kPa
35.7 mL, Tj = 293 K
P2 = 101.3 kPa, V2 = ?, T2 = 273 K
V2 = 31 mL
29. PV = nRT g/mol = 2.62 g/0.065512 mol
n=
a) 22 K = -2510C1 b) 756QC = 1029 K
T,
27. P, = 103 kPa, 22 L, Ti = 288 K
P1V1 - P2V2
P2 = 15 kPa, V2 = 108, T2 = ?
5 6.3gCa3(P04)2
23. [H+] = 10-ph = 10-™ = 3.16 x lO"8
-Li
(110 kPa)(1.46 L) = 40.0 g/mol
(8.31 kPa-L/K«mol)(295 K) =0.065512 mol
33. H2 + Br2 2HBr
2 mol H.,0
18.02 gH20
ir-©?- 1
31. C27H56 + 41 02-> 27 C02 + 28 H20
32.# mol = 1.37 g + 342.34 g/mol = 0.00400 mol
q = cmAT
= (4.18 J/goC)(2300 g)(2.370C) = 22785 J
J/moi= 22785 J + 0.00400 mol = 5694 kJ/mol
Bond kJ/mol
H-H
Br-Br
H-Br
436
193
366
#
1
1
required
# released
436
193
2
629
Ho + Bro-+ 2HBr + 103 kJ
732
732
103