Reactivity 2.1. How much? The amount of chemical change

How many moles of FeS2 are required to produce 64.07 g of SO₂?

4FeS₂ (s) + 11O₂ (g) → 2Fe₂O₃ (s) + 8SO₂ (g) 

A. 0.50 

B. 1.0 

C. 2.0 

D. 4.0 

Answer: A. 0.50

Mass of SO₂ = 64.07 g

Molar mass of SO₂ = 32.07 (S) + 2 × 16 (O) = 64.07 g/mol

Moles of SO₂ = `frac{64.07}{64.07}`= 1.0 mol

Moles of FeS₂ = `frac{1.0}{2}`= 0.5 mol

How many grams of potassium azide, KN_3, are needed to produce 68.1 dm3 of N₂ (g) at  STP? 

Molar gas volume at STP = 22.7 dm³ mol^-1; M of KN₃ = 81.13 g mol^-1. 2KN₃ (s) → 3N₂ (g) + 2K (s) 

A. 40.57 

B. 81.13 

C. 162.26 

D. 243.39 

Answer: C. 162.26 

Molar gas volume = 22.7 dm³ mol^-1

Moles of N₂ = `frac{68.1}{22.7}`= 3.00 mol 

Step 1. Use mole ratio between KN₃ and N₂

From equation: 

2 KN₃ → 3N₂

So: `frac{2}{3}`mol KN₃ produces 1 mol N₂

Thus: 

Moles of KN₃ = 3.00 × `frac{2}{3}`= 2.00 mol 

Step 2. Convert moles to grams 

Molar mass M(KN₃) = 81.13 g/mol 

Mass = 2.00 × 81.13 = 162.26 g 

0.8 dm³ of nitrogen monoxide is mixed with 0.6 dm³ of oxygen under standard  conditions. The reaction below takes place until one of the reactants is used up.

2NO (g) + O₂ (g) → 2NO₂ (g) 

What is the volume of the mixture under standard conditions after the reaction has taken place? 

A. 0.80 dm³. 

B. 1.00 dm³. 

C. 1.40 dm³. 

D. 1.20 dm³.

Answer: B. 1.00 dm³. 

From the equation, 2NO : 1O₂ 

→ For 0.8 dm³ NO, the required O₂  =`frac{0.8}{2}`= 0.4 dm³

We have 0.6 dm³ O₂ available → O₂ is in excess and NO is limiting.

Step 1. Calculate gases after reaction 

   • NO consumed: all 0.8 dm³ 

   • O₂ consumed: 0.4 dm³ 

   • O₂ left: 0.6 – 0.4 = 0.2 dm³ 

   • NO₂ produced: same volume ratio as NO (2NO → 2NO₂) → 0.8 dm³

Step 2. Total gas after reaction 

Vₜₒₜₐₗ = V_(NO₂) + V_{(O₂ left)} 

Vₜₒₜₐₗ = 0.8 + 0.2 = 1.0 dm³ 

0.10 mol of hydrochloric acid is mixed with 0.10 mol of magnesium carbonate.

2HCl (aq) + MgCO₃ (s) → MgCl₂ (aq) + H₂O (l) + CO₂ (g) 

Which is correct? 

Answer: D. 

2HCl + MgCO₃ → MgCl₂ + H₂O + CO₂ 

Given: 0.10 mol HCl and 0.10 mol MgCO₃ 

Needed HCl to consume all MgCO₃ = 2 × 0.10 = 0.20 mol → we only have 0.10 mol, so HCl is limiting. 

From the ratio 2HCl : 1CO₂, moles of CO₂ formed: 

`"n_(CO_2)` = `frac{0.10}{2}`= 0.05 mol 

What is the sum of all the coefficients (integers) when the equation below is balanced?

_(CH₃)₂NNH₂ + _N₂O₄ → _N₂ + _CO₂ + _H₂O 

A. 14 

B. 12 

C. 11 

D. 9 

Answer: B. 12. 

aC₂H₈N₂ + bN₂O₄ → cN₂ + dCO₂ + eH₂O 

   • Count N: 

     o Left: 2a + 2b 

     o Right: 2c 

→ c = a + b 

   • Count C: 

     o Left: 2a 

     o Right: d

→ d = 2a 

   • Count H: 

     o Left: 8a 

     o Right: 2e 

→ e = 4a 

   • Count O: 

     o Left: 4b 

     o Right: 2d + e = 2(2a) + 4a = 8a 

→ 4b = 8a 

→ b = 2a 

   • Substitute a = 1 (to keep smallest integers) 

Then b = 2, c = a + b = 3, d = 2a = 2, e = 4a = 4 

1(C₂H₈N₂) + 2(N₂O₄) → 3N₂ + 2CO₂ + 4H₂O 

   • Check atom balance 

   • Sum of coefficients 

1 + 2 + 3 + 2 + 4 = 12 

Nitrogen(II) oxide, NO, is made from the oxidation of ammonia, NH₃:

4NH₃ + 5O₂ → 4NO + 6H₂O 

An 8.52 g sample of NH₃ forms 14.75 g of NO. What is the percentage yield of NO?

A. 40%. 

B. 60%. 

C. 80%. 

D. 98%. 

Answer: D. 98%. 

4NH₃ + 5O₂ → 4NO + 6H₂O 

From the stoichiometry, NH₃ : NO = 1 : 1 in moles. 

   • Moles NH₃ =`frac{8.52 g}{17.03 "g" "mol^-1}` ≈ 0.500 mol 

   • Theoretical moles NO = 0.500 mol ⇒ theoretical mass = 0.500 × 30.01 ≈ 15.01 g

Percentage yield:

% yield = `frac{actual}{"theoretical"}*100 = frac{14.75}{15.01}*100`≈ 98% 

Potassium chlorate(VII) is prepared by three consecutive reactions:

Cl₂ (g) + 2KOH (aq) → KCl (aq) + KClO (aq) + H₂O (l)

3KClO (aq) → 2KCl (s) + KClO₃ (s) 

4KClO₃ (s) → 3KClO₄ (s) + KCl (s) 

If the overall percentage yield is 50%, what amount (mol) will be produced from 283.60 g of chlorine gas? 

A. 0.25 mol. 

B. 1.00 mol. 

C. 0.50 mol. 

D. 2.00 mol. 

Answer: C. 0.50 mol. 

1. Cl₂ + 2KOH → KCl + KClO → 1 mol Cl₂ gives 1 mol KClO 

2. 3KClO → 2KCl + KClO₃ → 1 mol KClO gives `frac{1}{3}`mol KClO₃ 

3. 4KClO₃ → 3KClO₄ + KCl → 13mol KClO₃ gives `frac{3}{4}*frac{1}{3}`= `frac{1}{4}`mol KClO₄.

So 1 mol Cl₂ ⇒14mol KClO₄ theoretically. 

Given 283.60 g of Cl₂: 

`n_(Cl_2)` =`frac{283.60}{70.90}` = 4.00 mol 

Theoretical KClO₄ = 4.00 ×14 = 1.00 mol 

Overall yield = 50% ⇒ actual = 1.00 × 0.50 = 0.50 mol 

Butyl ethanoate is an ester used as a flavouring. This ester can be synthesized from  butan-1-ol by two different processes. 

Process 1 is a one-step process that involves a reversible reaction: 

CH₃CH₂CH₂CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₂CH₂CH₃ + H₂O

6.25 g of butan-1-ol forms 6.57 g of butyl ethanoate. 

a. Calculate the percentage yield for process 1.  

b. Calculate the atom economy for process 1.  

Process 2 is a two-step process. Thionyl chloride (SOCl₂) is a volatile and reactive liquid used as a chlorinating agent which reacts with water to form sulfur dioxide and  hydrochloric acid. Ethanoyl chloride (CH₃COCl) hydrolyses with water to form  hydrochloric acid and ethanoic acid. 

CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl 

CH₃CH₂CH₂CH₂OH + CH₃COCl → CH₃COOCH₂CH₂CH₂CH₃ + HCl

5.450 g of ethanoic acid produces 9.806 g of butyl ethanoate. 

c. Calculate the overall percentage yield for process 2. 

d. Calculate the overall atom economy for process 2. 

e. Explain why process 2 has a high percentage yield but a low atom economy.

f. Suggest two reasons why butyl ethanoate is manufactured by process 1 rather than by process 2.

a. Percentage yield 

Molar mass of butan-1-ol = 74.12 g/mol 

Moles butan-1-ol = `frac{6.25}{74.12}`= 0.0843 mol

The reaction ratio is 1 : 1 → theoretical moles of ester = 0.0843 mol

Molar mass of butyl ethanoate = 116.16 g/mol 

Theoretical mass = 0.0843 × 116.16 = 9.79 g 

Percentage yield = `frac{6.57}{9.79}*100` = 67.1% 

b. Atom economy 

Atom economy = `frac{M_("desired product")}{ΣM_("reactants")}*100` 

Sum of reactants = 74.12 + 60.05 = 134.17 g/mol 

Desired product = 116.16 g/mol

Atom economy = `frac{116.16}{134.17}*100` = 86.6% 

c. Percentage yield 

M(ethanoic acid) = 60.05 g/mol 

Moles =`frac{5.450}{60.05}` = 0.0908 mol 

Theoretical ester = 0.0908 × 116.16 = 10.56 g 

Percentage yield = `frac{9.806){10.56}*100` = 92.9% 

d. Overall atom economy 

Total reactants (simplified overall): 

CH₃COOH + SOCl₂ + CH₃CH₂CH₂CH₂OH 

M = 60.05 + 118.97 + 74.12 = 253.14 g/mol 

Desired product = 116.16 g/mol 

Atom economy =1`frac{116.16}{253.15}*100`= 45.9% 

e. Explain why process 2 has high yield but low atom economy 

   • Process 2 is irreversible, giving a nearly complete conversion → high yield.

   • However, it produces many by-products (SO₂, HCl), so only a small portion of  atom mass ends up in the desired ester → low atom economy. 

f. Why process 1 is preferred 

   • Greener/safer: Avoids toxic SOCl₂ and corrosive HCl. 

   • Fewer steps: Cheaper, simpler, and more environmentally friendly (less waste). 

Phenol, C₆H₅OH, is an important pharmaceutical raw material for pharmaceutical  synthesis including aspirin. It is produced industrially by the cumene process which  involves the oxidation of benzene and propene. 

C₆H₆ + C₃H₆ + O₂ → C₆H₅OH + CH₃COCH₃ 

Another new method of producing phenol involves the use of solid zeolites as the  catalyst. The chemical equation for the process is shown below. This involves the  oxidation of benzene by nitrogen(I) oxide. 

C₆H₆ + N₂O → C₆H₅OH + N₂ 

a. Calculate the atom economy of the two processes. Hence state the process that is more  sustainable. 

b. Other than atom economy, state three other factors that are considered in determining  the sustainability and degree of green chemistry present in an industrial chemical  process. 

a. Atom economy 

Process 1 (cumene route): 

C₆H₆ + C₃H₆ + O₂ → C₆H₅OH + CH₃COCH₃ 

   • Mᵣ (reactants) = 78.11 + 42.08 + 32.00 = 152.19 g · mol^-1

   • Desired product (phenol) Mᵣ = 94.11 g · mol^-1

   • Atom economy = `frac{94.11}{152.19}*100`94.11 ≈ 61.8% 

Process 2 (zeolite, using N₂O): 

C₆H₆ + N₂O → C₆H₅OH + N₂ 

   • Mᵣ (reactants) = 78.11 + 44.01 = 122.12 g · mol^-1

   • Desired product = 94.11 g ·mol^-1

   • Atom economy =`frac{94.11}{122.12}*100`94.11 ≈ 77.0% 

→ The `frac{"zeolite"}{N_2O}` process is more sustainable (higher atom economy).

b. Three other sustainability/green-chemistry factors 

   • Overall yield & selectivity: High conversion to phenol with minimal by-products.

   • Energy efficiency: Low temperature/pressure; reduced heating/cooling demand.

   • Hazard/waste profile: Toxicity and environmental impact of reagents/by products/solvents (e.g., safe, recyclable catalysts; benign solvents or solvent-free  operation). 

Copper can exist in +1, +2 and +3 oxidation states in its compounds. After heating with  a Bunsen burner, 24.74 g of an unknown oxide of copper for 3 minutes left 19.76 g of  copper. 

a. Calculate the amounts (mol) of copper and oxygen. Use these values to determine the empirical formula of the oxide and state the oxidation state of the copper.

b. Suggest why the mass of solid obtained by heating 24.74 g of a copper oxide may be  greater than 19.76 g, giving one design improvement for your proposed suggestion. Ignore any possible errors in the weighing procedure. 

a. Empirical formula & oxidation state 

Mass of oxide heated = 24.74 g 

Mass of Cu after heating = 19.76 g 

→ Mass of O removed = 24.74 – 19.76 = 4.98 g 

Moles: 

   • Cu:`frac{19.76}{63.55}`≈ 0.311 mol 

   • O:`frac{4.98}{16.00}`4.98 ≈ 0.311 mol 

Ratio Cu : O ≈ 1 : 1 → empirical formula = CuO, so Cu is in the +2 oxidation state. b. Why the solid might weigh more than 19.76 g & design improvement During heating/cooling in air, copper (or partially reduced oxide) can re-oxidize (e.g.,  Cu → CuO), or incomplete reduction leaves some CuO, giving a final mass greater than  pure Cu. 

Design improvement:

Heat in a reducing / oxygen-free atmosphere (e.g., pass hydrogen or natural gas/CO over  the sample, use a covered crucible or tube furnace, and cool under inert gas or in a  desiccator) to prevent oxidation and ensure complete reduction.