Chem. 12 Review Test 1 Answers
Pick two formulas that match
each classification:
1. e b Acid
a) KOH e) HCl
2. c d Covalent
Nonacid b) CH3COOH f) NH4Cl
3. h f Salt
c) N2O g) Ba(OH)2
4. a g Base
d) HOH h) AgNO3
5. Calculate the molarity of the solution
formed when 200.0 g of NaCl is dissolved in
100. mL of H2O.
Molarity =
200g x 1 mole
58.5g = 34.2 M
0.100 L
6. How many grams of AgCl are required to
prepare 150.0 mL of 0.200 M solution?
.150L x 0.200 mole x 143.4 g =
4.30 g
1 L 1 mole
7. How many litres of 0.200 M AgCl are
needed to provide 50. g of AgCl?
50g x 1 mole x 1 L = 1.7 L
143.4g 0.200 mole
8. 100. g of AlCl3 is dissolved in 200.0 mL H2O, calculate [Al3+] and [Cl-].
100g x 1 mole
Molarity = 133.5 g
= 3.745 M AlCl3 ® Al3+ + 3Cl-
0.200 L
3.745 M 3.75 M
11.2M
9. In three runs of a titration 36.9, 34.4 and
34.3 mL of 0.200 M NaOH was required to neutralize a 25.0 mL sample of H2CO3. Calculate the molarity of
the acid.
H2SO4 + 2NaOH
® Na2SO4 + 2HOH
0.0250
L
0.3435 L
? M
0.200
M
[ H2SO4] = 0.03435 L x 0.200 mole x 1 mole H2SO4
1 L 2 mole NaOH
0.0250 L
= 0.137
M
10. Calculate the molarity of the excess NaOH in the solution formed
by mixing 50.0 mL of 0.200 M HCl with 50.0 mL of 0.200 M HNO3 and 75.0
mL of 0.300 M NaOH. Begin by writing a chemical equation for the reaction.
Assume that HCl will have the same effect on NaOH as HNO3 . Add the moles of HCl and HNO3 together.
HX + NaOH ® NaX + HOH
0.0500L HCl x 0.200
mole = 0.0100 mol 0 .0750 L x 0.300
mol = .0225 mole
1 L
1 L
0.0500L HNO3 x 0.200
mole = 0.0100 mol
1 L
Total Acid
I
0.0200 mole 0.0225 mole
C
0.0200 mole 0.0200 mole
E
0 mole 0.0025
mole
Total Volume =
175 mL = 0.175 L Molarity =
0.0025 mole =
0.014 M
0.175
L
11. A empty beaker has a mass
of 29.86 g. The same beaker is filled with 0.250 L with a solution of CaCl2
and weighs 87.26 g.
The solution is evaporated
to dryness and the mass of the beaker and solid is 62.31 g. Calculate the
molarity of the solution.
Mass of CaCl2
= 62.31 – 29.86 = 32.45g
Molarity =
32.45g x 1 mole
111.1g = 1.17 M
0.250 L
12. Complete the reaction equations.
i) Formula Equation/Chemical Equation
2AgNO3 (aq) + Na2SO4 (aq) ® Ag2SO4(s) + 2NaNO3(aq)
ii) Total Ionic Equation
2Ag+(aq) + 2NO3-
+ 2Na+(aq) +
SO42-
® Ag2SO4(s) + 2Na+(aq) + 2NO3-(aq)
iii) Net Ionic Equation
2Ag+(aq) +
SO42-
® Ag2SO4(s)
13.
Complete the formula equation:
2H3PO4(aq) +
3Sr(OH)2(aq)
® Sr3(PO4)2(s) +
6HOH(l)
Complete the complete ionic equation:
6H+(aq) + 2PO43- + 3Sr2+(aq) +
6OH- ® Sr3(PO4)2(s) + 6HOH(l)
Complete the net ionic equation:
6H+(aq) + 2PO43- + 3Sr2+(aq) +
6OH- ® Sr3(PO4)2(s) + 6HOH(l)
14.
Complete the formula equation:
Fe3(PO4)2(aq) +
3Zn(s) ® 3Fe(s) + Zn3(PO4)2(s)
Complete the complete ionic equation:
3Fe2+(aq) + 2PO43- + 3Zn(s) ® 3Fe(s) + Zn3(PO4)2(s)
Complete the net ionic equation:
3Fe2+(aq) + 2PO43- + 3Zn(s) ® 3Fe(s) + Zn3(PO4)2(s)
15. 40.0 mL of 0.600 M AlCl3 is added to 80.0 mL of
water. What are all ion concentrations?
AlCl3
®
Al3+ + 3Cl-
40.0 x 0.600 M = 0.200
M
0.200
M 0.600
M
120.0
16. 200.0 mL of 0.150 M AlCl3 is added to 200.0 mL 0.250
M BaCl2, calculate the [Ba2+], [Al3+] and the
[Cl-] immediately after mixing the two solutions.
AlCl3
®
Al3+ + 3Cl-
200.0 x 0.150 M = 0.0750
M
0.0750
M 0.225 M
400.0
BaCl2
®
Ba2+ + 2Cl-
200.0 x 0.250 M = 0.125
M
0.125
M 0.250
M
400.0
[Ba2+] = 0.125
M
[Al3+]
= 0.0750 M
[Cl-] = 0.225
M + 0.250
M = 0.475
M
17. A solution of 0.100 M H3C6H5O7(s) neutralized with 13.56 mL of
0.330 M KOH. Calculate the volume required.
H3C6H5O7 + 3KOH ® 1K3C6H5O7 + 3HOH
?g
0.01356
L
0.330 M
0.01356 L KOH x 0.330 moles x 1
mole H3C6H5O7 x 1 L = 0.0149 L H3C6H5O7
1
L 3 mole KOH 0.100 mole