To determine the amount of AIBr3 produced, we first need to identify the limiting reactant. This is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To do this, we need to calculate the amount of product that would be produced from each reactant separately and compare the two values.
From the balanced chemical equation, we can see that 2 moles of AI react with 3 moles of BR2 to produce 2 moles of AIBr3. Therefore, we need to convert the given masses of AI and BR2 to moles.
molar mass of AI = 27 g/mol
moles of AI = 54 g / 27 g/mol = 2 mol
molar mass of BR2 = 159.8 g/mol
moles of BR2 = 474 g / 159.8 g/mol = 2.97 mol
To calculate the amount of AIBr3 produced from each reactant, we use the mole ratio from the balanced chemical equation:
moles of AIBr3 produced from AI = 2 mol AI × (2 mol AIBr3 / 2 mol AI) = 2 mol AIBr3
moles of AIBr3 produced from BR2 = 2.97 mol BR2 × (2 mol AIBr3 / 3 mol BR2) = 1.98 mol AIBr3
Since the moles of AIBr3 produced from BR2 are less than the moles produced from AI, BR2 is the limiting reactant.
To calculate the mass of AIBr3 produced, we can use the mole ratio from the balanced chemical equation and the molar mass of AIBr3:
moles of AIBr3 produced = 1.98 mol
mass of AIBr3 produced = 1.98 mol × 266.7 g/mol = 529 g
Therefore, 529 grams of AIBr3 are produced.