Name: ___[ KEY ] ________________________________ Date: ________________ Period: ____ REVIEW Unit 4 Test (Chp 4) : Aqueous Reactions and Solution Stoichiometry Multiple Choice (20 questions) (50% of score) polyatomic ions (names, charges, formulas) 6 strong acids (what does it mean to say a “strong” “acid” ?) strong bases (group I and Ca, Ba, Sr with hydroxide, OH – ) (what is a “strong” “base” ?) electrolytes (strong, weak, non) (and what particles are actually in solution) ALWAYS soluble ions (Group I , NH 4 + , NO 3 – ) NET ionic equations (comp – diss – cross – net – bal) oxidation numbers of elements in compounds (like what is the ox. # of C in oxalate ion C 2 O 4 2– ?) redox reactions (LEO says GER) activity of metals (which metals are least active – or – which metals do not lose electrons as easily) gas forming reactions ( ____ reacted with ____ will form which gas: H 2 (g) or CO 2 (g) ) solution stoichiometry with Molarity ( L g ) ( g L ) ( g and L M ) ( dilutions M 1 V 1 = M 2 V 2 ) Free Response (32 points) (3 questions) (50% of score) 1
(NH4)2S has NH4+ which is soluble with all the listed anions,
and
(NH4)2S contains S2– which makes H2S(g) when H+ is added.
Comp: AgNO3(aq) + KI(aq) AgI(s) + KNO3(aq)
Diss: AgNO3(aq) + KI(aq) AgI(s) + KNO3(aq)
Cross: AgNO3(aq) + KI(aq) AgI(s) + KNO3(aq)
K+ and NO3– are spectator ions that do not change charge or state.
Double Replacement Rxns show all products for B-E are soluble containing ALWAYS soluble ions.
A) KOH(aq) + Cu(NO3)2(aq) KNO3(aq) + Cu(OH)2(s)
Cu(OH)2 precipitates
E) ALL NH4+ compounds are soluble
Pb(NO3)2 (aq)
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BaI2(aq) + Na2SO4(aq) BaSO4(s) + NaI(aq)
E) BaSO4 precipitates, but Na+ and I– do not change charge or state
H2SO4(aq) + NaOH(aq) H2O(l) + Na2SO4(aq)
C) strong acid plus strong base produces a dissolved salt (Na2SO4) not in the net ionic equation(Na+ and SO4
2– are crossed out as spectator ions)
Strong Base = completely dissociate in water (all hydroxide compounds of group I and CBS from group II) with OH–
HNO3 is strong KCl is a salt, NOT and acid (no H)
A) True b/c NH4+ has an extra H+ it can donate as an acid (though not a strong acid)
and NH3 can accept an H+ acting like a base (though not a strong base)
What is/are the spectator ion(s) in the reaction between BaI2(aq) and Na2SO4(aq)?
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C) for example:
CaCO3 + H2SO4 CaSO4 + H2O + CO2
OR
NaHCO3 + HCl NaCl + H2O + CO2
B) for example:Mg + 2 HCl MgCl2 + H2
D) Ag cannot replace H+ from acids (not active enough) (below hydrogen in activity series)so it must be below Ni as well in the same series.
D) Cu cannot replace H+ from acids (not active enough) (below hydrogen in activity series)
E) S has a +6 in the sulfate ion in K2SO4
2K + S + 4O = 02(+1) + S + 4(–2) = 0
+2 + S + –8 = 0 +2 + S = +8
S = +6
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C) Cl goes from a +5 in chlorate in KClO3 to a –1 as chloride in KCl
(K+ does not change)
(O gets oxidized from –2 in chlorate in KClO3 to 0 in O2)
A) H goes from a +1 in water to a 0 as hydrogen gas (still a +1 in NaOH)
C) Zn goes from a 0 to a +2while Ag goes from +1 to 0All others are double- replacement rxns
D) 0.20 mol BaI2 x 2 mol I – = 0.40 mol I–
1 mol BaI2
OR BaI2(aq) Ba2+(aq) + 2 I–(aq)
0.20 mol 0.20 mol 0.40 mol
Ba2+ Ba2+I– I– I– I–
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1.00g BeO x 1 mol BeO = 0.040 mol BeO 0.040 mol = 0.20 M25 g BeO 0.200 L
You should do the math with simple numbers, then move decimal places in your answers:
100/25 = 4 0.04 4/2 = 2 0.02 0.2(move left 2 b/c 1.00 not 100) (move left 2 b/c .04 not 4, right 1 b/c div. by .2 not 2)
M1V1 = M2V2 (key word dilution, though it won’t always be used)
(0.500)(40.0) = M2(200) (leave in mL is fine)
20.0 = M2 M2 = 0.100 M 200
0.500 L CH3OH x 0.500 mol CH3OH x 32.0 g CH3OH = 8.00 g CH3OH 1 L CH3OH 1 mol CH3OH
You should do the math with simple numbers, then move decimal places in your answers:
5/5 = 25 0.250 0.250 x 32 = ¼ of 32 = 8 (move left 2 b/c 0.5 & 0.5 not 5 & 5)
How many grams of methanol, CH3OH (MW = 32.0 g∙mol–1) are there in 500. mL of a 0.500 M CH3OH solution?
10.0g HClO4 x 1 mol HClO4 = 0.100 mol HClO4 0.100 mol = 0.40 M 100 g HClO4 0.250 L
.
.
27. Beaker X and beaker Y each contain 1.0 L of solution, as shown above. A student combines the solutions by pouring them into a larger, previously empty beaker Z and observes the formation of a white precipitate.Assuming that volumes are additive, which of the following sets of solutions could be represented by the diagram above?
Beaker X Beaker Y Beaker Z(A) 2.0 M AgNO3 2.0 M MgCl2 4.0 M Mg(NO3)2 and AgCl(s)(B) 2.0 M AgNO3 2.0 M MgCl2 2.0 M Mg(NO3)2 and AgCl(s)(C) 2.0 M AgNO3 1.0 M MgCl2 1.0 M Mg(NO3)2 and AgCl(s)(D) 2.0 M AgNO3 1.0 M MgCl2 0.50 M Mg(NO3)2 and AgCl(s)
Section II Free ResponseCalculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS. It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do not. Attention should be paid to significant figures.
The concentration of Mg2+ in Y is half the concentration of Ag+ in X.
The concentration of Mg2+ in Z is half of what it is in Y because the volume doubled.
Best answer choice must represent these ratios:
(D) 2.0 M AgNO3 1.0 M MgCl2 0.50 M Mg(NO3)2
1. A student is assigned the task of preparing 50. mL of 6 M HNO3 to use in an experiment.
(a) The student is provided with a stock solution of 16 M HNO3 , two 100 mL graduated cylinders that can be read to ±1 mL, a 100 mL beaker that can be read to ±10 mL, safety goggles, rubber gloves, a glass stirring rod, a dropper, and distilled H2O.
(i) Calculate the volume, in mL, of 16 M HNO3 that the student should use for preparing 50. mL of 6 M HNO3 . (1)
(ii) Briefly list the steps of an appropriate and safe procedure for preparing the 50. mL of 6 M HNO3. Only materials selected from those provided to the student (listed above) may be used. (2)
(iii) Explain why it is not necessary to use a volumetric flask (calibrated to 50.00 mL ±0.05 mL) to perform the dilution. (1)
*(The “ones” digit is called the “units” digit properly)
In a second experiment, a student is given 2.94 g of a mixture containing anhydrous MgCl2 and KNO3. To determine the percentage by mass of MgCl2 in the mixture, the student uses excess AgNO3(aq) to precipitate the chloride ion as AgCl(s).
(b) Starting with the 2.94 g sample of the mixture dissolved in water, briefly describe the steps necessary to quantitatively determine the mass of the AgCl precipitate. (2)
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M1V1 = M2V2
(16 M)(V1) = (6 M)(50. mL)
V1= 19 mL or 20 mL (to one significant digit)
1) Wear safety goggles and rubber gloves.
2) Then measure 20 mL of 16M HNO3 using a 100 mL graduated cylinder.
3) Measure 30 mL of distilled H2O using a 100 mL graduated cylinder.
4) Transfer the water to a 100 mL beaker.
5) Add the acid to the water with stirring.
(AAA : “always add acid” to water (never water to acid) to minimize exothermic reaction and splashing)
1 pt for any 3
2 pts for all 5
The graduated cylinders provide sufficient precision in volume measurement to provide one significant digit to the tens digit for 20 mL (OR two significant digits to the units* digit for 19 mL) as required for the lab, making the use of the volumetric flask unnecessary.
(c) The AgCl precipitate is filtered, washed, dried, and weighed to constant mass in a filter crucible. The student determines the mass of the AgCl precipitate to be 5.48 g. On the basis of this information, calculate each of the following.
(i) The number of moles of MgCl2 in the original mixture (2)
(ii) The percent by mass of MgCl2 in the original mixture (1)
2. The identity of an unknown solid is to be determined. The compound is one of the seven salts in the following table.
KNO3 Ba(OH)2•8H2O CaCO3 CuSO4•5H2O
NaOH BaSO4 Ni(NO3)2•6H2O
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Add excess AgNO3 .
- Separate the AgCl precipitate (by filtration).
- Wash the precipitate and dry the precipitate completely.
- Determine the mass of AgCl by difference.
5.48 g AgCl × 1 mol AgCl = 0.0382 mol AgCl 143.32 g AgCl
1.82 g MgCl2 × 100% = 61.9% MgCl2 by mass2.94 g sample
Use the results of the following observations or laboratory tests to explain how each compound in the table may be eliminated or confirmed. The tests are done in sequence from (a) through (e).
(a) The unknown compound is white. In the table below, cross out the two compounds that can be eliminated using this observation. Be sure to cross out these same two compounds in the tables in parts (b), (c), and (d). (2)
KNO3 Ba(OH)2•8H2O CaCO3 CuSO4•5H2O
NaOH BaSO4 Ni(NO3)2•6H2O
(b) When the unknown compound is added to water, it dissolves readily. In the table below, cross out the two compounds that can be eliminated using this test. Be sure to cross out these same two compounds in the tables in parts (c) and (d). (2)
KNO3 Ba(OH)2•8H2O CaCO3 CuSO4•5H2O
NaOH BaSO4 Ni(NO3)2•6H2O
(c) When AgNO3(aq) is added to an aqueous solution of the unknown compound, a white precipitate forms. In the table below, cross out one compound that can be eliminated using this test. Be sure to cross out the same compound in the table in part (d). (1)
KNO3 Ba(OH)2•8H2O CaCO3 CuSO4•5H2O
NaOH BaSO4 Ni(NO3)2•6H2O
(d) When the unknown compound is carefully heated, it loses mass. In the table below, cross out one compound that can be eliminated using this test. (1)
KNO3 Ba(OH)2•8H2O CaCO3 CuSO4•5H2O
NaOH BaSO4 Ni(NO3)2•6H2O
(e) Describe a test that can be used to confirm the identity of the unknown compound identified in
part (d). Limit your confirmation test to a reaction between an aqueous solution of the unknown compound and an aqueous solution of one of the other soluble salts listed in the tables above. Describe the expected results of the test; include the formula(s) of any product(s).
3. For each of the following three reactions, in part (i) write a balanced equation for the reaction with coefficients in terms of lowest whole numbers, and in part (ii) answer the question about the reaction. Assume that solutions are aqueous unless otherwise indicated. Represent substances in
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(blue, seen in lab)
Mix an aqueous solution of Ba(OH)2•8H2O with an aqueous solution of CuSO4∙ 5H2O .
Ba(OH)2(aq) + CuSO4(aq) BaSO4(s) + Cu(OH)2(s)
If the unknown is Ba(OH)2 , then a precipitate will form. (actually two precipitates, BaSO4 and Cu(OH)2 )
NaCl + AgNO3 will form AgCl(s)Ba(OH)2 + AgNO3 will formAgOH(s)
Al(NO3)3 will not
(transition metals have d orbitals for e– to move in emitting colors (freq’s) of light)
(both CaCO3 & BaSO4 are insoluble precipitates)
solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space for scratch work, but only equations that are written in the answer boxes provided will be graded.
(a) A solution of barium hydroxide is added to a solution of acetic acid (ethanoic acid).
(i) Balanced equation: (4)
(ii) Which substance is a weak electrolyte. Explain. (1)
___ HC 2H3O2 is a weak electrolyte because it exists mostly as molecules in solution__
rather than as dissociated ions______________________________________________
Ba(OH)2 + HC2H3O2 H2O + Ba(C2H3O2)2
or Ba(OH)2 + CH3COOH H2O + Ba(CH3COO)2
(b) A solution of potassium sulfide is added to a solution of iron(III) chloride.
(i) Balanced equation: (4)
(ii) Classify this reaction in two ways. (1)
___double-replacement and precipitation_____________________________________
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