Answer:
3. The mass of ethanol required is approximately 0.522869 g
The mass of ethanoic acid required is approximately 0.68156 g
4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes
5. The mass of silver nitrate required is approximately 14.53 grams
6. The mass of copper oxide that would be needed is approximately 31.86 grams
7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams
b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams
c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams
Explanation:
3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate
The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles
∴ The number of moles of ethanol = 1/88.11 moles
The number of moles of ethanoic acid = 1/88.11 moles
The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g
The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g
4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron
The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles
The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes
5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles
The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams
6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles
The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams
7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles
2 moles of NaOH produces 1 mole of Zn(OH)₂
0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂
The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams
The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams
b. 6 moles of NaOH produces 2 moles Al(OH)₃
20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃
The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams
c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;
20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂
The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams
