Chemistry assignment - hess\'s law, moles PDF

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Chemistry assignment - hess's law, moles, thermochemistry and much more highschool based review questions....

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60. The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3% Al and 79.7% chlorine. Determine the empirical and molecular formulas of aluminium chloride. 79.70 20.3 𝑎𝑛𝑑 𝐶𝑙 : 27 35.5 267 𝐴𝑙𝐶𝑙3 = = 2 = 𝐴𝑙 2 𝐶𝑙6 133.5 𝐴𝑙:

61.

Sodium reacts with water as follows. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H 2(g)

1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to 250 cm3. Calculate the concentration, in mol dm–3, of the resulting sodium hydroxide solution. 1.15 = 0.05 23 𝑁𝑎 = 0.05 𝑚𝑜𝑙𝑒𝑠 0.05 = 0.20 𝑚𝑜𝑙 𝑑𝑚−3 0.25 (Total 3 marks)

62.

(i)

Calcium carbonate is added to separate solutions of hydrochloric acid and ethanoic acid of the same concentration. State one similarity and one difference in the observations you could make.

One similarity I believe would be in the observations is that a source of fizz would be found. A difference would be that the HCl solution in the reaction is more vigorous. (2)

(ii)

Write an equation for the reaction between hydrochloric acid and calcium carbonate. 2HCl + CaCO3 → CaCl2 + CO2 + H20

(iii) Determine the volume of 1.50 mol dm–3 hydrochloric acid that would react with exactly 1.25 g of calcium carbonate. 𝐶𝑎𝐶𝑂3 =

1.25 = 0.012 100.09

2x 0.012 = 0.025 number of moles for HCl 0.025 = 0.017𝑑𝑚−3 1.50

(3)

(iv)

Calculate the volume of carbon dioxide, measured at 273 K and1.01×10 5 Pa, which would be produced when 1.25 g of calcium carbonate reacts completely with the hydrochloric acid.

V= n/RT/ovf R= 8.31 0.0125x273x8.314/1.01x105 = 2.8x10−4 (2) (Total 9 marks

63.

An organic compound, A, containing only the elements carbon, hydrogen and oxygen was analysed. (a)

A was found to contain 54.5 C and 9.1 H by mass, the remainder being oxygen. Determine the empirical formula of the compound. 𝑐𝑎𝑟𝑏𝑜𝑛 =

54.5 12

ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 𝑜𝑥𝑦𝑔𝑒𝑛 =

9.1 1.01

36.4 16

= 𝐶2 𝐻4 0 (3)

(b)

A 0.230 g sample of A, when vaporized, had a volume of 0.0785 dm3 at 95C and 102 kPa. Determine the relative molecular mass of A.

M x RT/PV 95 degrees = 368K 0.230 𝑥 8.31 𝑥 368 = 87.8 102 𝑥 0.0785 𝐴 = 87.8 (c)

Determine the molecular formula of A using your answers from parts (a) and (b).

A = 87.8 Given/ worked C2H4O = 2 x 12.01 + 4 x 1.01 + 16 = 44.06 = 87.8/44.06 = 2 = 𝐶4 𝐻8 02

(1)(Total 7 marks)

64.

An organic compound A contains 62.0 by mass of carbon, 24.1 by mass of nitrogen, the remainder being hydrogen. (i) 𝐶= 𝑁= 𝐻=

Determine the percentage by mass of hydrogen and the empirical formula of A. 62.0 12

= 5.16

24.1 = 14 13.9 1.01

1.72

= 13.8

5.16/1.72 = 3 1.72/1.72 = 1 13.8/1.72 = 8 = C3NH8

(ii)

Define the term relative molecular mass. 1

The term relative molecular mass is the weighted average mass of an atom compared to12 the mass of the atom carbon 12. (2)

(iii)

The relative molecular mass of A is 116. Determine the molecular formula of A.

Given/worked 166 =2 58 = 𝐶6 𝑁2 𝐻16 (1) (Total 6 marks)

65.

Same as question number 64

An organic compound A contains 62.0 by mass of carbon, 24.1 by mass of nitrogen, the remainder being hydrogen. (i)

Determine the percentage by mass of hydrogen and the empirical formula of A. (3)

(ii)

Define the term relative molecular mass. ................................................................................................................................... ...................................................................................................................................

................................................................................................................................... (2)

(iii)

The relative molecular mass of A is 116. Determine the molecular formula of A. ................................................................................................................................... ................................................................................................................................... (1) (Total 6 marks)

66.

Propane and oxygen react according to the following equation. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Calculate the volume of carbon dioxide and water vapor produced and the volume of oxygen remaining, when 20.0 dm3 of propane reacts with 120.0 dm3 of oxygen. All gas volumes are measured at the same temperature and pressure. 20.0𝑑𝑚 3 𝑥

3𝑑𝑚3 𝐶𝑜2 = 60.0𝑑𝑚 3 = 𝐶𝑂2 1𝑑𝑚3 𝐶3 𝐻8

20.0𝑑𝑚3 𝑥

4𝑑𝑚3 𝐻20 = 80.0𝑑𝑚3 = 𝐻20 1𝑑𝑚3 𝐶3𝐻8

20.0𝑑𝑚 3 𝑥

5𝑑𝑚3 𝑂2 = 100.0 𝑑𝑚3 = 𝑂2 1𝑑𝑚3 𝐶3𝐻8

120.0-100 = 20.0𝑑𝑚3 𝑜𝑥𝑦𝑔𝑒𝑛 𝑟𝑒𝑚𝑎𝑛𝑖𝑛𝑔 Total 3 marks)

67.

State and explain what would happen to the pressure of a given mass of gas when its absolute temperature and volume are both doubled.

When the temperature doubles the pressure will as well double, however when the volume doubles the pressure halves. (Total 3 marks)

68.

(i)

Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxide and 0.657 g of water when it undergoes complete combustion.

Crocetin = 1.00g CO2 = 2.68g

44.01 12

H20 = 0.657g

18 2

=

=

2.68 ?

0.657 ?

= 0.73 of carbon = 0.073 of hydrogen

O2 = 1 – 0.73 – 0.073 = 0.197 of oxygen 0.197/16 = 0.0123

0.0609 /0.0123 = 5 0.073/0.0123 = 6 0.0123/0.0123 = 1 = C5H6O

(ii)

Determine the molecular formula of crocetin given that 0.300 mole of crocetin has a mass of 98.5 g

Given/worked C5H60 5(12.01) x 6(1.01) x 16.00 = 82.11g 98.5/0.300 = 328g 328g/82.11g = 4 = C20 H24 O4 (2) (Total 8 marks)

A solution containing ammonia requires 25.0 cm3 of 0.100 mol dm–3 hydrochloric acid to reach the equivalence point of a titration.

69.

(i)

Write an equation for the reaction of ammonia with hydrochloric acid

NH3 + HCl = NH4 Cl (1)

(ii)

Calculate the amount (in mol) of hydrochloric acid and ammonia that react.

0.0250 𝑥 0.100𝑚𝑜𝑙 = 0.00250𝑚𝑜𝑙 (2)

(iii)

Calculate the mass of ammonia in the solution.

0.0250 𝑥 17.04𝑚𝑜𝑙 = 0.0426𝑔 (2) = (Total 5 marks)

A toxic gas, A, consists of 53.8% nitrogen and 46.2% carbon by mass. At 273 K and1.01×105 Pa, 1.048 g of A occupies 462 cm3. Determine the empirical formula of A.Calculate the molar mass of the compound and determine its molecular structure.

70.

N= C=

53.8 14.01 46.2 12.01

= 3.85 𝑚𝑜𝑙 = 3.85 mol

= CN M = mrt/pv

=

1.048𝑥8.31𝑥273 101𝑥0.462𝑑𝑚 3

= 51.0g

(Total 3 marks)

71.

An oxide of copper was reduced in a stream of hydrogen as shown below

After heating, the stream of hydrogen gas was maintained until the apparatus had cooled. The following results were obtained. Mass of empty dish = 13.80 g Mass of dish and contents before heating = 21.75 g Mass of dish and contents after heating and leaving to cool = 20.15 g (a)

Explain why the stream of hydrogen gas was maintained until the apparatus cooled. It was maintained to determine and see that copper did not react with oxygen. (1)

(b)

Calculate the empirical formula of the oxide of copper using the data above, assuming complete reduction of the oxide. 21.75-20.15 = 1.60g of oxygen 20.15 – 13.80 = 6.35g of copper 6.35 x 1/63.55g = 0.1 mol of copper 1.60 x 1/16g = 0.1 mol of oxygen = CuO (3)

(c)

Write an equation for the reaction that occurred. Cu O + H2 → Cu + H20 (1)

(d)

State two changes that would be observed inside the tube as it was heated.

One change can be so that the color changes and the second change can be the vaporization of water. (2) (Total 7 marks)

72.

Copper metal may be produced by the reaction of copper(I) oxide and copper(I) sulfide according to the below equation. 2Cu2O + Cu2S → 6Cu + SO 2 A mixture of 10.0 kg of copper(I) oxide and 5.00 kg of copper(I) sulfide was heated until no further reaction occurred. (a)

Determine the limiting reagent in this reaction, showing your working. 10000 x 1/143.1g = 69.8 5000 x 1/159.16 = 31.4

69.8/2 = 34.9 31.4/1 = 31.4 Cu2 S = the limiting (3)

(b)

Calculate the maximum mass of copper that could be obtained from these masses of reactants. 31.4/1 = x/6 = 188.4 x 63.55 = 11972.8/1000 11.97 = 12.0 (2) (Total 5 marks)

73.

The reaction below represents the reduction of iron ore to produce iron. 2Fe2O3 + 3C → 4Fe + 3CO2

A mixture of 30 kg of Fe 2O3 and 5.0 kg of C was heated until no further reaction occurred. Calculate the maximum mass of iron that can be obtained from these masses of reactants. Fe2O3 → C→

5

30 2 𝑥 55.845+3 𝑥 16

= 0.187𝑚𝑜𝑙

= 0.416 𝑚𝑜𝑙

12 0.187

Fe → 𝑥 4 = 0.374 𝑚𝑜𝑙 2 0.374 x 55.845 = 21kg 21 kg is the maximum mass of iron that can be obtained (Total 5 marks)

74.

0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution, BaCl2(aq) is added to precipitate all the sulfate ions as barium sulfate, BaSO 4(s). The precipitate is filtered and dried and weighs 0.672 g. (a)

Calculate the amount (in mol) of barium sulfate formed. 137.34 + 32.06 + 4(16) = 233.4 0.672/233.4 = 0.002879 (2)

(b)

Determine the amount (in mol) of the alkali metal sulfate present. 0.002879

c) Determine the molar mass of the alkali metal sulfate and state its units. 0.502/0.002879 = 174.3 (2)

(d)

Deduce the identity of the alkali metal, showing your workings. 174-(32+4x16)/2 = 39

(2)

(e)

Write an equation for the precipitation reaction, including state symbols. K2SO4 + BaCl2 → BaSO4 + 2KCl (2) (Total 9 marks)

75.

0.600 mol of aluminium hydroxide is mixed with 0.600 mol of sulfuric acid, and the following reaction occurs: 2Al(OH) 3(s) + 3H 2SO4(aq) → Al2(SO4)3(aq) + 6H2O(l) (a)

Determine the limiting reactant. 0.600/2 = 0.3 0.600/3 = 0.3 = H2SO4 is the limiting u (2)

(b)

Calculate the mass of Al2(SO4)3 produced. 0.600 x 1/3 x 342/1 = 68.4 (2)

(c)

Determine the amount (in mol) of excess reactant that remains. 0.600 x 2/3 = 0.400 0.600 – 0.400 = 0.200mol (1)

(d)

Define a Brønsted-Lowry acid and a Lewis base. Brønsted-Lowry acid It is a substance that donates a proton or an (H+) hydrogen ion. Lewis base It is a compound that donates and electron pair to an acceptor compound. (2)

(e)

H2SO4(aq) is a strong acid. State the name and the formula of any weak acid. CH3 COOH – acetic acid (1)

(Total 8 marks)...