Answer:
Therefore, approximately 8.5 mL of chlorine gas can be produced from the decomposition of 25 g of aluminum chloride at 28°C and 750 mmHg.
Step-by-step explanation:
The decomposition of aluminum chloride will not produce oxygen. Instead, it produces aluminum oxide and chlorine gas. The balanced chemical equation for this reaction is:
2AlCl3(s) → Al2O3(s) + 3Cl2(g)
To determine the volume of chlorine gas produced from the decomposition of 25 g of aluminum chloride, we need to use the ideal gas law, which is PV = nRT, where P is the pressure of the gas in atmospheres, V is the volume of the gas in liters, n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature of the gas in Kelvin.
First, we need to determine the number of moles of aluminum chloride. The molar mass of AlCl3 is 133.34 g/mol, so:
25 g AlCl3 × (1 mol AlCl3/133.34 g AlCl3) = 0.187 moles AlCl3
According to the balanced chemical equation, 2 moles of AlCl3 will produce 3 moles of Cl2 gas. Therefore, 0.187 moles of AlCl3 will produce:
0.187 moles AlCl3 × (3 moles Cl2/2 moles AlCl3) = 0.2805 moles Cl2
Now we can use the ideal gas law to calculate the volume of Cl2 gas produced. We need to convert the temperature from Celsius to Kelvin by adding 273.15.
PV = nRT
V = nRT/P
V = (0.2805 mol) × (0.0821 L·atm/mol·K) × (301.15 K) / (750 mmHg × 1 atm/760 mmHg)
V = 0.0085 L or 8.5 mL (rounded to two significant figures)
Therefore, approximately 8.5 mL of chlorine gas can be produced from the decomposition of 25 g of aluminum chloride at 28°C and 750 mmHg.