Answer:
Explanation:
To find out how many grams of aluminum chloride (AlCl3) form when 0.25 moles of chlorine gas (Cl2) react, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the reaction between chlorine gas (Cl2) and aluminum (Al) to form aluminum chloride (AlCl3) is:
2 Al + 3 Cl2 -> 2 AlCl3
From the balanced equation, we can see that 3 moles of chlorine gas (Cl2) are required to react with 2 moles of aluminum (Al) to produce 2 moles of aluminum chloride (AlCl3).
Given that we have 0.25 moles of Cl2, we can use the stoichiometry of the reaction to find out how many moles of AlCl3 will form:
0.25 moles Cl2 * (2 moles AlCl3 / 3 moles Cl2) = 0.25 * (2/3) moles AlCl3
Now, let's calculate the number of moles of AlCl3:
0.25 * (2/3) = 0.16667 moles AlCl3
Now that we know the number of moles of AlCl3 formed, we can use the molar mass of AlCl3 to convert moles to grams:
Molar mass of AlCl3 = 133.22 g/mol
Grams of AlCl3 = moles of AlCl3 * molar mass of AlCl3
Grams of AlCl3 = 0.16667 moles * 133.22 g/mol ≈ 22.203 grams
So, approximately 22.203 grams of aluminum chloride (AlCl3) form when 0.25 moles of chlorine gas (Cl2) react.
