Please help solve;
2Al +3Cl2 --> 2AlCl3
You are give 31.0g Al and 36.0g of Cl2.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0 g of aluminum?
Here is an example of a stoichiometry problem I've posted. Just follow the procedure shown to work the second part of your question. For the first part, see how much AlCl3 (with excess Al) can be produced with 36.0 g Cl2, compare with the answer from the second part. The value for both, where one functions as a limiting reagent, will be the smaller of the two.
http://www.jiskha.com/science/chemistry/stoichiometry.html
figure the moles you have of aluminum, and chlorine.
Start with the aluminum. IF you have more than 3/2 of that of chlorine, then aluminum is the limiting reageant. If you have less than 3/2 of chlorine, chlorine is the limiiting reageant.
Take the mole of the limiting regeant, and use the mole ratio in the balanced equation to gind moles of aluminum cloride.
Example: aluminum is the limiting regent. If you have 3.4 moles of aluminum, you will get 2/2 * 3.4 moles of aluminum chloride.
Example: chlorine is the limiting regent. If you have 4.5 moles of chlorine, then you get 4.5*2/3 moles of aluminum cloride.
To find the number of moles of aluminum chloride that can be produced from 31.0 g of aluminum, we need to use the given balanced chemical equation:
2Al + 3Cl2 -> 2AlCl3
First, we need to determine the number of moles of aluminum (Al) present in 31.0 g. We can do this by using the molar mass of aluminum, which is 26.98 g/mol:
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 31.0 g / 26.98 g/mol
Calculating this, we get:
Moles of Al = 1.15 mol
Next, we can use the stoichiometry of the balanced equation to determine the moles of aluminum chloride (AlCl3) that can be produced from the moles of aluminum.
From the balanced equation, we can see that:
2 moles of Al produces 2 moles of AlCl3
Therefore, the 1.15 mol of Al can produce the same number of moles of AlCl3:
Moles of AlCl3 = Moles of Al = 1.15 mol
Therefore, if you had excess chlorine, you could produce 1.15 moles of aluminum chloride (AlCl3) from 31.0 g of aluminum.
To find the number of moles of aluminum chloride that can be produced from 31.0 g of aluminum, you need to follow a few steps:
Step 1: Determine the molar mass of aluminum (Al)
The molar mass of aluminum (Al) is 26.98 g/mol.
Step 2: Calculate the number of moles of aluminum (Al)
To calculate the number of moles, divide the given mass (31.0 g) by the molar mass of aluminum:
moles of Al = mass of Al / molar mass of Al
moles of Al = 31.0 g / 26.98 g/mol
Step 3: Determine the stoichiometry of the balanced equation
From the balanced equation, 2Al + 3Cl2 -> 2AlCl3, you can see that the ratio of aluminum (Al) to aluminum chloride (AlCl3) is 2:2, or 1:1. This means that for every 2 moles of aluminum reacting, 2 moles of aluminum chloride are produced.
Step 4: Calculate the number of moles of aluminum chloride (AlCl3)
Since the ratio is 1:1, the number of moles of aluminum chloride will be the same as the number of moles of aluminum:
moles of AlCl3 = moles of Al = 31.0 g / 26.98 g/mol
Using this calculation, you will find the number of moles of aluminum chloride that can be produced from 31.0 g of aluminum.