What is the maximum mass in grams of aluminum chloride that could be obtained from 2.50 mol of barium chloride and excess aluminum sulfate? this is the balanced equation for the reaction: al2(so4)3 + 3bacl2--> 3baso4 + 2alcl3. bacl2 molar mass is 208.3 g/mol and alcl3molar mass is 133.3 g/mol.

a. 521 g
b. 334 g
c. 222 g
d. 55.8

The maximum amount of aluminium chloride that could be obtained is 222 grams (option C)

Explanation:

Step 1: Data given

Number of moles barium chloride = 2.50 mol

Molar mass of BaCl2 = 208.3 g/mol

Molar mass of AlCl3 = 133.3 g/mol

Step 2: The balanced equation:

Al2(SO4)3 + 3BaCl2 → 3BaSO4 + 2AlCl3

Step 3: Calculate moles of AlCl3

For 1 mol aluminium sulfate we need 3 moles Barium chloride to produce 3 moles of barium sulfate and 2 moles of Aluminium chloride

For 2.50 mol of BaCl2 we have (2/3) * 2.50 = 1.667 moles of aluminium chloride

Step 4: Calculate mass of aluminium chloride

Mass AlCl3 = moles AlCl3 * molar mass AlCl3

Mass AlCl3 = 1.667 moles * 133.3 g/mol

Mass AlCl3 = 222 grams

The maximum amount of aluminium chloride that could be obtained is 222 grams (option C)