What is the maximum mass of aluminum chloride that can be formed when reacting 30.0 g of aluminum with 35.0 g of chlorine?
To find the maximum mass of aluminum chloride that can be formed, we need to determine the limiting reactant in the given chemical equation. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by writing a balanced chemical equation for the reaction between aluminum and chlorine to form aluminum chloride:
2Al + 3Cl₂ → 2AlCl₃
According to the balanced equation, the stoichiometric ratio between aluminum (Al) and aluminum chloride (AlCl₃) is 2:2. In other words, 2 moles of aluminum react with 2 moles of aluminum chloride.
First, we need to convert the given masses of aluminum and chlorine into moles. To do this, we will use the molar mass of each element:
Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of chlorine (Cl₂) = 35.45 g/mol
Converting the masses to moles:
Moles of aluminum = 30.0 g ÷ 26.98 g/mol = 1.1128 mol
Moles of chlorine = 35.0 g ÷ 35.45 g/mol = 0.9869 mol
Next, we compare the moles of aluminum (1.1128 mol) to the stoichiometric ratio in the balanced equation (2 moles of Al: 2 moles of AlCl₃). Since the ratio is 1:1, the moles of aluminum chloride formed will also be 1.1128 mol.
Finally, we need to find the mass of aluminum chloride formed by multiplying the moles by its molar mass:
Molar mass of aluminum chloride (AlCl₃) = (26.98 g/mol) + 3(35.45 g/mol) = 133.34 g/mol
Mass of aluminum chloride = 1.1128 mol × 133.34 g/mol = 148.3 g
Therefore, the maximum mass of aluminum chloride that can be formed when reacting 30.0 g of aluminum with 35.0 g of chlorine is 148.3 g.
65.80
I will be happy to critique your thinking.
aluminium is excess.
let x be the maximum mass of AlCl3
(x/133.5)/(35/71)=2/3
x=
(find x by yourself)