Based on the equation, how many grams of br2 are required to react completely with 42.3 grams of alcl3? alcl3 + br2 → albr3 + cl2 66.5 grams 71.2 grams 76.1 grams 80.2 grams

76.1 grams.

Explanation:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,

2.0 moles of AlCl₃ reacts with 3.0 moles of Br₂ to produce 2.0 moles of AlBr₃ and 3.0 moles of Cl₂.

no. of moles of AlCl₃ = mass/molar mass = (42.3 g)/(133.34 g/mol) = 0.3172 mol.

Using cross multiplication:

2.0 moles of AlCl₃ react completely with → 3.0 moles of Br₂.

0.3172 mol of AlCl₃ react completely with → ??? moles of Br₂.

∴ The no. of moles of Br₂ = (0.3172 mol)(3.0 mol)/(2.0 mol) = 0.4759 mol.

∴ The amount of grams of Br₂ are required to react completely with 42.3 grams of AlCl₃ = (no. of moles of Br₂)(molar mass) = (0.4759 mol)(159.81 g/mol) = 76.05 g ≅ 76.1 g.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa