Answer:
1- 16.0 L.
2- 189.0 g.
3- 95.0 g.
Explanation:
1. What volume of O₂ at STP is produced from the reaction of 212 grams of Au₂O₃?
- To solve this problem, we use the balanced equation:
2Au₂O₃ → 4Au + 3O₂
- It is clear that 2.0 moles of Au₂O₃ are decomposed to give 4.0 moles of Au and 3.0 moles of O₂.
- Firstly, we should convert the grams of Au₂O₃ (212.0 g) reacted into no. of moles using the relation: n = mass / molar mass.
- The no. of moles of Au₂O₃ = mass / molar mass = (212.0 g) / (441.93 g/mol) = 0.479 mol.
Using cross multiplication:
2.0 moles of Au₂O₃ produces → 3.0 moles of O₂, from the balanced equation.
0.479 mole of Au₂O₃ produces → ??? moles of O₂.
- ∴ The no. of moles of O₂ produced from the decomposition of 212.0 g of Au₂O₃ = (0.479 mol)(3.0 mol) / (2.0 mol) = 0.72 mol.
- Finally, we can get the volume of O₂ produced using the gas law: PV = nRT.
where, P is the pressure of the gas (P = 1.0 atm at STP),
V is the volume of the gas in L (V = ??? L),
n is the no. of moles of the gas (n = 0.72 mol),
R id the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas (T = 273.0 K at STP).
- ∴ The volume of O₂ produced = nRT/P = (0.72 mol)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm) = 16.1 L ≅ 16.0 L.
2. How many grams of Au are produced?
By the same way used in Q1,
Using cross multiplication:
2.0 moles of Au₂O₃ produces → 4.0 moles of Au, from the balanced equation.
0.479 mole of Au₂O₃ produces → ??? moles of Au.
- ∴ The no. of moles of Au produced from the decomposition of 212.0 g of Au₂O₃ = (0.479 mol)(4.0 mol) / (2.0 mol) = 0.958 mol.
- Finally, we can get the no. of grams of Au produced using the law: mass = n x atomic mass.
- The no. of grams of Au produced = n x atomic mass = (0.958 mol)(196.96 g/mol) = 188.69 g ≅ 189.0 g.
3. How much Au₂O₃ is needed to form 85 grams gold?
- By the same way and using the stichiometry of the balanced equation:
- Firstly, we need to get the no. of moles of Au produced (85.0 g) using the law: n = mass / atomic mass.
- The no. of moles of Au produced n = mass / atomic mass = (85.0 g) / (196.96 g/mol) = 0.43 mol.
Using cross multiplication:
2.0 moles of Au₂O₃ produces → 4.0 moles of Au, from the balanced equation.
??? mole of Au₂O₃ produces → 0.43 moles of Au.
∴ The no. of moles of Au₂O₃ needed to from 85.0 g of Au = (0.43 mol)(2.0 mol) / (4.0 mol) = 0.215 mol.
- Finally, we can get the no. of grams of Au₂O₃ needed using the law: mass = n x atomic mass.
- The no. of grams of Au₂O₃ needed = n x atomic mass = (0.215 mol)(441.93 g/mol) = 95.35 g ≅ 95.0 g.
