Final answer:
To determine the mass of aluminum chloride required, we need to use the stoichiometry of the reaction. The mass of aluminum chloride required is 271.52 g.
Step-by-step explanation:
To determine the mass of aluminum chloride required, we need to use the stoichiometry of the reaction. The balanced equation is: 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s). From the equation, we can see that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. The molar mass of sodium is 22.99 g/mol and the molar mass of aluminum chloride is 133.34 g/mol.
Using the given mass of sodium (90g) and the molar mass of sodium, we can calculate the number of moles of sodium. Then, using the stoichiometry of the reaction, we can determine the number of moles of aluminum chloride. Finally, multiplying the number of moles of aluminum chloride by its molar mass will give us the mass of aluminum chloride required.
Therefore, the mass of aluminum chloride required is [(90g / 22.99 g/mol) * (2 mol AlCl3 / 2 mol Al) * (133.34 g/mol AlCl3)] = 271.52 g.