End-of-chapter exam1~3

Part I: Multiple Choices (2’×10=20’)

Choose the best answer, and put A, B, C or D on the single line beside the number of each question. Each question hasONLY ONEbest answer.

___B___1.What is the balanced equation for the reaction betweentrimanganese tetroxide and aluminium?

A.Mn3O4+ 4Al → 2Al2O3+ 3MnB.3Mn3O4+ 8Al → 4Al2O3+ 9Mn

C.2Mn3O4+ 8Al → 4Al2O3+ 6MnD.4Mn3O4+ 8Al → 4Al2O3+ 12Mn

___D___2.What is the Avogadro constant?

A.Thenumber of atoms in 1 gram of any element

B.Thenumber of atoms in 1 mole of molecules of anyelement

C.Thenumber of atoms in the formula mass of acompound

D.Thenumber of atoms in 1 mole of atoms of anyelement

___C___3.What is the mass of carbon dioxide (molar mass =44.0 gmol−1) in 1200 cm3of carbon dioxide?

A.2200 gB.52.8 gC.2.2 gD.0.05 g

1.2 L / 24 Lmol-1* 44.0 gmol-1

___D___4.What is the concentration of chloride ions in a solutioncontaining 0.02 mol of calcium chloride, CaCl2, in 200 cm3of solution?

A.0.01 mol dm−3B.0.02 mol dm−3C0.1 mol dm−3D0.2 mol dm−3

___B___ 5.What information is given by a molecular formula?

A.Theratio of each type of atom in a molecule

B.Thenumber of each type of atom in a molecule

C.Thearrangement of the atoms in a molecule

D.Thenumber of atoms in a molecule

___D___ 6.Theequation for a reaction is:Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s).What is the ionic equation for the reaction?

A.Ba2+(aq) + Na2SO4(aq) → 2Na+(aq) + BaSO4(s)

B.Ba(NO3)2(aq) + 2Na+(aq) → 2NaNO3(aq) + Ba2+(s)

C.Ba2+(aq) + SO42−(aq) → Ba2+(s) + SO42−(aq)

D.Ba2+(aq) + SO42−(aq) → BaSO4(s)

___A___ 7.An oxide of copper contains 0.635 g of copper and 0.080 g ofoxygen by mass.(Ar values: Cu = 63.5, O = 16.0)

What is the empirical formula of this oxide of copper?

A.Cu2OB.CuOC.CuO2DCu2O3

___A___ 8.100 cm3of an aqueous solution of sodium hydroxide ofconcentration 0.10 mol dm−3reacts with 200 cm3of aqueousphosphoric acid. Theequation for the reaction is:2NaOH + H3PO4→ Na2HPO4+ 2H2O

What is the concentration of the phosphoric acid?

A.0.025 mol dm−3B.0.050 mol dm−3C.0.10 mol dm−3D.0.040 mol dm−3

___B___ 9.Which one of the following reactions is accompanied by thelargest percentage increase in volume?

A.4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)B.2NH3(g) → N2(g) + 3H2(g)

C.CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)D.S(s) + O2(g) → SO2(g)

___C___ 10.Hydrochloric acid reacts with barium hydroxide:2HCl + Ba(OH)2→ BaCl2+ 2H2O.What is the volume of hydrochloric acid of concentration0.050 mol dm−3which exactly neutralises 25.0 cm3of a0.10 mol dm−3solution of barium hydroxide?

A.25 cm3B.50 cm3C.100 cm3D.200 cm3

Part II: Short Responses (5’×4=20’)

Answer the questions briefly with two or three sentences.

1. What do you understand by the term “relative atomic mass (Ar)”? Why doesn’t it have a unit of “gram”?

mass ratio (1) of an atom (1) against the 1/12 (1) of the mass of C-12 (1)

because it’s a ratio, unit of gram is cancelled(1)

2. The mass spectrum of neon has three peaks:20Ne (90.9%),21Ne (0.3%) and22Ne (8.8%). Calculate Arfor neon.

Ar = 20×90.9% + 21×0.3% + 22×8.8% (3)

= 18.18 + 0.063 + 1.936 = 20.179 = 20.2 (round to 3 significant figures) (1)

3. The following is the range of color change for phenolphthalein. In our titration, when it changes color, does it mean pH = 7? Or should it be higher or lower than 7.0? Give your reason.

Indicator

pH < 8.2

pH = 8.2 – 10.0

Phenolphthalein

colorless

purple-red

No (1), it will be higher than 7 (1).

When we begin titrating NaOH(aq) with HCl(aq), the pH decreases from a high point(1). And the ending point is not 7.0, according to the form (1). Because the ending point is 8.2, higher than 7.0(1), the ultimate pH for the solution should be higher than 7.0.

4. Write down the following ionic equations. If this is a precipitation reaction, put a “√” in the brackets; if not, leave it blank.

a. HCl(aq) + AgNO3(aq)b. Ba(OH)2(aq) + CuSO4(aq)c. Na2CO3(aq) + HCl(aq)

Ag+(aq) + Cl-(aq)→AgCl(s)1’

________________________________________________________________________(√1’ )

Ba2+(aq) + SO42-(aq) + Cu2+(aq) + 2OH-(aq)→BaSO4(s) + Cu(OH)�2(s)1’

________________________________________________________________________(√1’ )

CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)1’

________________________________________________________________________(    )

Part III: Long Questions (60’)

1. Calcium carbonate (CaCO3) is usually used for preparation of CO2in laboratories.

a. Calculate the maximum mass of CO2that can be produced with 20.0 g CaCO3.[5]

b. Write out the ionic equation for this reaction, labeling the state for every substance?[2]

c. Point out the spectator ion for the last equation. What do you understand by the term “spectator ion”?[3]

d. If 20.0 g CaCO3reacts with excess hydrochloric acid of 40.0 mL.Calculate the concentration of the CaCl2solution after the reaction goes completely.(Ignore the volume change of the solution)[5]

(Ar values: Ca = 40.0, C = 12.0, O = 16.0, Cl = 35.5)

Total = 15

a. Mr(CaCO3) = 40.0 + 12.0 + 48.0 = 100Mr(CO2) = 12.0 + 32.0 = 44.02’

CaCO3→CO2/ CaCO3+ 2HCl→CaCl2+ H2O + CO21’

100.044.0

20.0 gm1’

m = 20.0×44 / 100 = 8.8 g1’

b. CaCO3(s) + 2H+(aq)→Ca2+(aq) + CO2(g) + H2O(l)

correct equation 1’, correct labels 1’

c. spectator ions: Cl-(1)

Spectator ions are those which don’t actually take part in the reaction (1); and they can be cancelled in the ionic equation (1).

d. n(CaCO3) = 20.0 g / 100.0 gmol-1= 0.2 mol1’

CaCO3→CaCl2/ CaCO3+ 2HCl→CaCl2+ H2O + CO21’

11

0.2 mol0.2 mol1’

c = n/V1’

= 0.2 mol / 0.04 L = 5.00 mol/L1’

2. Naming compounds is a basic skill A-Level students should acquire.

a. Name the following inorganic and organic compounds: CO2, Fe(OH)3, C2H6, CH3CH2OH.[4]

b. Write out the corresponding molecular formulae for the following compounds:[4]

i. ammonium nitrateii. lithium chlorideiii. potassium hydrocarbonateiv. Carbon disulfide

c. Draw 3 conclusions about how to name an inorganic compound.[3]

d. Explain how to deduce the chemical formula of copper(II) sulfate.[4].

Total = 15

a. carbon dioxide, iron(III) hydroxide, ethane, ethanol (1’per each)

b. NH4NO3, LiCl, KHCO3, CS2(1’per each)

c. 1) name of the cation is put in front of that of the anion;1’

2) name of the cation stays the same as the element or the group; 1’

3) if the anion is simple, the ending of its name will be changed into–ide; 1’

One more choice: if the anion is complex, the name stays the same as the group

d. The ions presents: Cu2+SO42-2’

For electrical neutrality, we need one Cu2+ion for one SO42-ion.1’

So the formula is CuSO41’

3. Zinc is constantly used for producing hydrogen gas in labs.

a. Write out the fully labeled ionic equation for zinc’s reaction with hydrochloric acid. Which is the spectator ion?[3]

b. If 6.5 g Zn reacts with excessive hydrochloric acid, calculate the maximum volume of hydrogen being produced at r.t.p. (at r.t.p. 1 mole of gas occupies 24.0 L)[6]

c. Instead of Zn, Mg can also react with hydrochloric acid. So which metal of the same mass can produce more hydrogen? Explain.[6]

(Ar value: Zn = 65.0, H = 1.0, Mg = 24.0)Total = 15

a. Zn(s) + 2H+(aq)→Zn2+(aq) + H2(g) (correct equation 1’, correct labels 1’), spectator ion: Cl-1’

b. n(Zn) = m/M1’

= 6.5 g/ 65 gmol-1= 0.1 mol1’

Zn→H21’

1 mol24.0 L

0.1 molV1’

V = 24.0×0.1 / 11’

= 2.4 L1’

c. Zn→H21’Mg→H21’

65 g    2 g1’24 g2 g1’

From it we can see to produce the same mass of hydrogen (2 g), a larger mass of Zn (65 g) is needed than Mg (24 g).(1) Thus if the mass are the same for the two metals, Zn can produce less hydrogen (1).

4. There is not only one formula for one compound.

a. Fill in the form.[4]

b. The composition by mass of a hydrocarbon is 85.7% carbon and 14.3% hydrogen. Deduce the empirical formula for this compound.[6]

c. The complete combustion of a hydrocarbon with a mass of 14.0 g gives out 44.0 g CO2and 18.0 g H2O. Deduce the stoichiometry of the equation and tell the molecular formula of the hydrocarbon.[5]

(Arvalue: C = 12.0, H = 1.0, O = 16.0)Total = 15

a. H2O2, H—O—O—H, HOOH, HO

C2H6,

, CH3CH3, CH3(0.5’for each)

b.CH1’

85.7%14.3%   1’

n:85.7% / 12.0 = 0.071414.3% / 1 = 0.1432’

n(C) : n (H) = 0.0714 / 0.1431’

= 1 : 2   1’

c. Suppose the molecule formula for this hydrocarbon is CxHy, then

CxHy+ O2→CO2+H2O1’

44.0 g18.0 g

n:44.0/44.0 = 1 mol18.0 / 18.0 = 1 mol1’

stoichiometry:1:11’

Thus CxHy+ O2→1 CO2+ 1 H2O. According to the law of conservation of mass, x = 1, y = 2.1’

So the formula is CH2.1’