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.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.