Question

Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3? 2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2 65.2 grams 69.8 grams 72.1 grams 76.5 grams

Answer 1

65.2 grams

Step-by-step explanation:

We can first find the moles of AlCl3:

36.2 g AlCl3 / 133.34 g/mol = 0.27 moles AlCl3

Knowing the molar ratio of AlCl3 to Br2 in the reaction, we can calculate the number of moles of Br2 required:

0.27 moles AlCl3 * 3 moles Br2 / 2 moles AlCl3 = 0.405 moles Br2

Finally, we can convert the number of moles to grams:

0.405 moles * 159.808 g/mol = 65.02 g Br2

So, 65.02 grams of Br2 are required to react completely with 36.2 grams of AlCl3.