Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?

2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2

65.0 g of Br2 are required to react completely with 36.2 grams of AlCl3.

Explanation:

To get the amount of grams of Br2 that are required to react completely with 36.2 grams of AlCl3, we must first get the amount of moles of AlCl3 in 36.2 grams of AlCl3. We do this by dividing by the molar mass.

36.2 g : 133.34 g/mol = 0.271 mol

Now that we have the moles of AlCl3 we can get the moles of Br2 by multiplying by the mole ratio from the chemical equation.

0.271 mol AlCl3 * 3/2 = 0.407 mol Br2

Now that we have to moles of Br2, we can get the mass by multiplying by the molar mass.

0.407 mol * 159.81 g/mol = 65.0 g Br2

65.0 g of Br2 are required to react completely with 36.2 grams of AlCl3.