Answer:
The Minimum volume of chlorine gas required to completely react with 7.07 g of aluminum is 28.6 liters.
Step-by-step explanation:
The balanced chemical equation for the reaction between aluminum and chlorine gas is:
2Al + 3Cl2 → 2AlCl3
From the balanced equation, we can see that 2 moles of Al react with 3 moles of Cl2.
First, we need to convert the mass of Al to moles. The molar mass of Al is 26.98 g/mol.
Moles of Al = 7.07 g / 26.98 g/mol = 0.262 moles
Using the stoichiometry of the reaction, we find that the moles of Cl2 needed is:
Moles of Cl2 = 0.262 moles Al x (3 moles Cl2 / 2 moles Al) = 0.393 moles Cl2
Next, we use the ideal gas law (PV=nRT) to find the volume of Cl2. We need to convert the pressure to atmospheres first:
260 mmHg x (1 atm / 760 mmHg) = 0.342 atm
Then we can solve for V:
V = nRT / P = (0.393 moles)(0.0821 L·atm/mol·K)(298 K) / 0.342 atm = 28.6 L
So, the minimum volume of chlorine gas required to completely react with 7.07 g of aluminum is 28.6 liters.