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Amount Of Substance Flashcards

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1
Q

Nitroglycerine, C3H5N3O9, is an explosive which, on detonation, decomposes rapidly to form a large number of gaseous molecules. The equation for this decomposition is given below.

4C3H5N3O9( l ) → 12CO2( g ) + 10H2O( g ) + 6N2( g ) + O2( g )

A sample of nitroglycerine was detonated and produced 0.350 g of oxygen gas.

State what is meant by the term one mole of molecules.

A
  • Avogardro’s number of molecules
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2
Q

Calculate the number of moles of oxygen gas produced in this reaction, and hence deduce the total number of moles of gas formed.

( Mass of oxygen = 0.350 )
( Mr of oxygen = 16 )
( 4C3H5N3O9( l ) → 12CO2( g ) + 10H2O( g ) + 6N2( g ) + O2( g ) )

A

Moles of oxygen:

  • 0.35 / 32
  • = 7 / 640 = 1.09 x 10^-2

Total moles of gas:

  • Total number of moles of gas = 29
  • Total moles of gas = 29 x 7 / 640
  • = 203 / 640 = 0.317
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3
Q

Calculate the number of moles, and the mass, of nitroglycerine detonated.

( 4C3H5N3O9( l ) → 12CO2( g ) + 10H2O( g ) + 6N2( g ) + O2( g ) )
( Moles of oxygen = 7 / 640 )

A

Moles of nitroglycerine:

  • 4 x 7 / 640
  • = 7 / 160 = 0.0438

Mass of nitroglycerine:

  • 7 / 160 x 227
  • = 1589 / 160 = 9.93 g
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4
Q

A second sample of nitroglycerine was placed in a strong sealed container and detonated.
The volume of this container was 1.00 × 10^-3 m^3.
The resulting decomposition produced a total of 0.873 mol of gaseous products at a temperature of 1100 K.

State the ideal gas equation and use it to calculate the pressure in the container after detonation.
( The gas constant R = 8.31 J K^-1 mol^-1 )

A

Ideal gas equation:

  • pV = nRT

Pressure:

  • p = nRT / V
  • p = ( 0.873 ) ( 8.31 ) ( 1100 ) / ( 1 x 10^-3 )
  • p = 798009.3
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5
Q

Sodium chlorate( V ), NaClO3, contains 21.6% by mass of sodium, 33.3% by mass of chlorine and 45.1% by mass of oxygen

Use the above data to show that the empirical formula of sodium chlorate( V ) is NaClO3

A

Na:

  • Mass = 0.216
  • Mr = 23.0
  • Moles = 0.216 / 23
  • = 27 / 2875 = 9.39 x 10^-3

Cl:

  • Mass = 0.333
  • Mr = 35.5
  • Moles = 0.333 / 35.5
  • = 333 / 35500 = 9.38 x 10^-3

O:

  • Mass = 0.451
  • Mr = 16
  • Moles = 0.451 / 16
  • = 451 / 16000 = 0.0281

Empirical Na:

  • 9.39 x 10^-3 / 9.38 x 10^-3
  • = 852 / 851 = 1.00
  • Empirical Cl
  • 9.38 x 10^-3 / 9.38 x 10^-3
  • = 1

Empirical O:

  • 0.0281 / 9.38 x 10^-3
  • = 32021 / 10656 = 3.00
  • Hence empirical formula = NaClO3
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6
Q 
Sodium chlorate( V ) may be prepared by passing chlorine into hot aqueous sodium hydroxide. 
Balance the equation for this reaction below.

…. Cl2 + …. NaOH - > …. NaCl + NaClO3 + 3H2O

A
  • 3Cl2 + 6NaOH - > 5NaCl + NaClO3 + 3H2O
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7
Q

Potassium nitrate, KNO3, decomposes on strong heating, forming oxygen and solid Y as the only products.

A 1.00 g sample of KNO3 ( Mr = 101.1 ) was heated strongly until fully decomposed into Y.

Calculate the number of moles of KNO3 in the 1.00 g sample.

A
  • Moles = 1 / 101.1

- = 10 / 1011 = 9.89 x 10^-3

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8
Q

At 298 K and 100 kPa, the oxygen gas produced in this decomposition occupied a volume of 1.22 × 10^-4 m^3.
State the ideal gas equation and use it to calculate the number of moles of oxygen produced in this decomposition.
( The gas constant R = 8.31 J K^-1 mol^-1 )

A

Ideal Gas equation:

  • pV = nRT

Moles of oxygen:

  • n = pV / RT
  • n = ( 100 x 10^3 ) ( 1.22 x 10^-4 ) / ( 8.31 ) ( 298 )
  • n = 610 / 123819 = 4.93 x 10^-3
  • n = 4.93 x 10^-9
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9
Q

Compound Y contains 45.9% of potassium and 16.5% of nitrogen by mass, the remainder being oxygen.

State what is meant by the term empirical formula.

A
  • Simpliest ratio of atoms of each element in a compound
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10
Q

Use the data above to calculate the empirical formula of Y.

Compound Y contains 45.9% of potassium and 16.5% of nitrogen by mass, the remainder being oxygen.
( KNO3 )

A

K:

  • Mass = 0.459
  • Mr = 39.1
  • Moles = 0.459 / 39.1
  • = 27 / 2300 = 0.0117

N:

  • Mass = 0.165
  • Mr = 14.0
  • Moles = 33 / 2800 = 0.0118

O:

  • Mass = 0.376
  • Mr = 16
  • Moles = 47 / 2000 = 0.0235

Empirical K:

  • 0.0117 / 0.0117 = 1

Empirical N:

  • 0.0118 / 0.0117
  • = 253 / 252 = 1.00

Empirical O:

  • 0.0235 / 0.0117
  • = 1081 / 540
  • = 2.00
  • Hence empirical formula = KNO2
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11
Q

Deduce an equation for the decomposition of KNO3 into Y and oxygen.

( Compound Y = KNO2 )

A
  • 2KNO3 - > 2KNO2 + O2
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12
Q

The equation for the reaction between magnesium hydroxide and hydrochloric acid is shown below.

Mg( OH )2( s ) + 2HCl( aq ) → MgCl2( aq ) + 2H2O( l )

Calculate the volume, in cm^3, of 1.00 mol dm^-3 hydrochloric acid required to react completely with 1.00 g of magnesium hydroxide.

A

Moles Mg( OH )2:

  • Mr Mg( OH )2 = 58.3
  • Mass Mg( OH )2 = 1.00 g
  • Moles = 1 / 58.3
  • = 10 / 583 = 0.0172
  • Moles HCl = 2 x 0.0172
  • ( 2 : 1 )
  • = 20 / 583 = 0.0343
  • Volume HCl = 0.0343 / 1
  • = 0.0343
  • 0.0343 x 1000 = 34.3 cm^3
  • ( dm^3 > cm^3 )
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13
Q

When aluminium is added to an aqueous solution of copper( II ) chloride, CuCl2, copper metal and aluminium chloride, AlCl3, are formed.
Write an equation to represent this reaction.

A
  • 2Al + 3CuCl2 - > 2AlCl3 + 3Cu
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14
Q

Lead( II ) nitrate may be produced by the reaction between nitric acid and lead( II ) oxide as shown by the equation below.

PbO + 2HNO3 → Pb( NO3 )2 + H2O

An excess of lead( II ) oxide was allowed to react with 175 cm^3 of 1.50 mol dm^-3 nitric acid.
Calculate the maximum mass of lead( II ) nitrate which could be obtained from this reaction.

A

Moles HNO3:

  • 1.50 x 175 x 10^-3
  • = 0.2625

Moles Pb( NO3 )2:

  • 0.2625 / 2
  • = 0.131

Mr of Pb( NO3 )2:

  • = 331.2

Mass of Pb( NO3 )2:

  • 0.131 x 331.2
  • = 43.5g
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15
Q

Ammonia, NH3, reacts with sodium to form sodium amide, NaNH2, and hydrogen.

Write an equation for the reaction between ammonia and sodium.

A
  • 2Na + 2NH3 - > 2NaNH2 + H2
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16
Q

A hydrocarbon, W, contains 92.3% carbon by mass.
The relative molecular mass of W is 78.0.

Calculate the empirical formula of W.

A

C:

  • 92.3 / 12
  • = 923 / 120
  • = 7.69

H:

  • 7.7 / 1
  • = 7.7

C:

  • ( 923 / 120 ) / 7.7
  • = 923 / 924
  • = 1.0

H:

  • 7.7 / 7.7
  • = 1.0
  • Empirical formula = CH
17
Q

Calculate the molecular formula of W.

Empirical formula = CH
( Molecular mass of W = 78 )

A
  • Mr of empirical formula = 12 + 1
  • = 13
  • 78 / 13 = 6
  • Molecular formula = C6H6
  • ( Multiply every atom by 6 )
18
Q

One isotope of sodium has a relative mass of 23.

Calculate the mass, in grams, of a single atom of this isotope of sodium.
( The Avogadro constant, L, is 6.023 × 10^23 mol^-1 )

A
  • 23 / 6.023 x 10^23

- = 3.82 x 10^-23

19
Q

Titanium( IV ) chloride reacted with water as shown in the following equation.

TiCl4( l ) + 2H2O( l ) → 4HCl( aq ) + TiO2( s )

The reaction produced 200 cm^3 of a 1.20M solution of hydrochloric acid.
Calculate the number of moles of HCl in the solution and use your answer to find the original mass of TiCl4.

A

Moles HCl:

  • 1.20 x 200 x 10^-3
  • = 6 / 25
  • = 0.24

Mass of TiCl4:

  • Moles TiCl4 = 0.24 / 4
  • = 3 / 50
  • = 0.06
  • Mass = 0.06 x 190
  • = 57 / 5
  • = 11.4
20
Q

Calculate the volume of 1.10 M sodium hydroxide solution which would be required to
neutralise a 100 cm^3 portion of the 1.20 M solution of hydrochloric acid.

( Titanium( IV ) chloride reacted with water as shown in the following equation.

TiCl4( l ) + 2H2O( l ) → 4HCl( aq ) + TiO2( s )

The reaction produced 200 cm^3 of a 1.20M solution of hydrochloric acid. )

( Moles HCl = 0.24 in 200 cm^3 )

A
  • NaOH + HCl - > NaCl + H2O
  • Moles NaOH = Moles HCl
  • = 0.12 in 100cm^3
  • Volume = Moles / Concentration
  • Volume = 0.12 / 1.10
  • = 0.1090909091 dm^3
  • = 109 cm^3
21
Q

An excess of magnesium metal was added to a 100 cm^3 portion of the 1.20 M solution of hydrochloric acid.
Calculate the volume of hydrogen gas produced at 98 kPa and 20°C.

Mg( s ) + 2HCl( aq )→ 4 MgCl2( aq ) + H2( g )

( Titanium( IV ) chloride reacted with water as shown in the following equation.

TiCl4( l ) + 2H2O( l ) → 4HCl( aq ) + TiO2( s )

The reaction produced 200 cm^3 of a 1.20M solution of hydrochloric acid. )

( Moles of HCl in 100 cm^3 = 0.12 )

A
  • Moles of H2 = 0.06
  • ( 0.12 / 2 )
  • PV = nRT
  • V = nRT / P
  • V = ( 0.06 ) ( 8.31 ) ( 20 + 273 ) / ( 98 x 10^3 )
  • V = 1.490712245 x 10^-3 m^3
  • = 1.49 x 10^-3
22
Q

Calculate the maximum mass of sodium sulphide that can be obtained from 10.0 g of sulphur.

A
  • Sodium sulphide = Na2S
  • Moles S = 10 / 32.1
  • = 100 / 321
  • = 0.312
  • Mr Na2S = 78.1
  • Mass Na2S = ( 100 / 321 ) x 78.1
  • = 7810 / 321
  • = 24.3 g
23
Q

Calculate the minimum volume of hydrogen, in cm^3, at 298 K and 101.3 kPa, that is needed to form 5.00 g of hydrogen sulphide.

A
  • Hydrogen sulphide = H2S
  • Mr of Hydrogen sulphide = 34.1
  • Moles of hydrogen sulphide = 5 / 34.1
  • = 50 / 341
  • = 0.147
  • PV = nRT
  • V = nRT / P
  • V = ( 0.147 ) ( 8.31 ) ( 298 ) / ( 101.3 x 10^3 )
  • V = 3.59356229 x 10^-3 m^3
  • V = 3590 cm^3
24
Q

Q15

A

Chemistry (13 decks)

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