How many moles of aluminum chloride ALCL3 for when 81 g AL reacts with plenty of chlorine? (AI: 27 g/mol) 2AI + 3CI2 —> 2AICI3
The Correct Answer and Explanation is :
To determine how many moles of aluminum chloride (AlCl₃) are produced when 81 grams of aluminum (Al) reacts with plenty of chlorine gas (Cl₂), we first need to start with the balanced chemical equation:
[
2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3
]
Step 1: Calculate the number of moles of aluminum (Al)
The molar mass of aluminum (Al) is given as 27 g/mol. To find the number of moles of aluminum, we use the formula:
[
\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}}
]
Substituting in the given values:
[
\text{moles of Al} = \frac{81 \, \text{g}}{27 \, \text{g/mol}} = 3 \, \text{moles of Al}
]
Step 2: Use the stoichiometry of the reaction
From the balanced equation, we see that 2 moles of Al react to produce 2 moles of AlCl₃. This means that the molar ratio of Al to AlCl₃ is 1:1. Therefore, for every mole of aluminum, one mole of aluminum chloride is produced.
Since we have calculated that there are 3 moles of aluminum, we can conclude that:
[
\text{moles of AlCl}_3 = 3 \, \text{moles of Al}
]
Thus, when 81 grams of aluminum reacts with plenty of chlorine gas, it produces 3 moles of aluminum chloride (AlCl₃).
Conclusion:
In summary, starting with 81 grams of aluminum, we first calculated the moles of aluminum using its molar mass, finding that it yields 3 moles of Al. Based on the stoichiometry of the reaction, this results in the formation of 3 moles of AlCl₃. The balanced equation allows us to directly correlate the amount of reactant to the product, ensuring that we understand the precise quantities involved in the chemical reaction.