If you had excess chlorine, how many moles of of aluminum chloride could be produced from 20.0 g of aluminum?

If you had excess aluminum, how many moles of aluminum chloride could be produced from 25.0 g of chlorine gas Cl2?

To find the number of moles of aluminum chloride produced, we need to use the balanced chemical equation for the reaction between aluminum and chlorine to form aluminum chloride.

2 Al + 3 Cl2 → 2 AlCl3

1. If you had excess chlorine and wanted to determine the number of moles of aluminum chloride produced from 20.0 g of aluminum, you would follow these steps:

Step 1: Convert the mass of aluminum to moles.
The molar mass of aluminum (Al) is 26.98 g/mol.
molar mass of aluminum = 26.98 g/mol

moles of aluminum = mass (g) / molar mass (g/mol)
moles of aluminum = 20.0 g / 26.98 g/mol

Step 2: Use the stoichiometric ratio from the balanced equation to determine the moles of aluminum chloride produced.
From the balanced equation, we can see that 2 moles of aluminum reacts with 2 moles of aluminum chloride.

moles of aluminum chloride = moles of aluminum

2 Al = 2 moles of AlCl3

moles of aluminum chloride = 20.0 g / 26.98 g/mol

2. If you had excess aluminum and wanted to determine the number of moles of aluminum chloride produced from 25.0 g of chlorine gas (Cl2), you would follow these steps:

Step 1: Convert the mass of chlorine gas to moles.
The molar mass of chlorine gas (Cl2) is 70.90 g/mol.
molar mass of chlorine gas = 70.90 g/mol

moles of chlorine gas = mass (g) / molar mass (g/mol)
moles of chlorine gas = 25.0 g / 70.90 g/mol

Step 2: Use the stoichiometric ratio from the balanced equation to determine the moles of aluminum chloride produced.
From the balanced equation, we can see that 3 moles of chlorine gas reacts with 2 moles of aluminum chloride.

moles of aluminum chloride = (3/2) * moles of chlorine gas

moles of aluminum chloride = (3/2) * (25.0 g / 70.90 g/mol)

Please note that in both cases, it is assumed that the reactions proceed to completion.

To find the number of moles of aluminum chloride that can be produced from a given amount of aluminum or chlorine gas, we need to use the concept of moles and balanced chemical equations.

First, let's start with the reaction between aluminum (Al) and chlorine (Cl2) to form aluminum chloride (AlCl3):

2 Al + 3 Cl2 → 2 AlCl3

Now, let's solve each question step by step:

1) If you have excess chlorine, how many moles of aluminum chloride could be produced from 20.0 g of aluminum?

Step 1: Convert the mass of aluminum (Al) to moles.
To do this, we need to know the molar mass of aluminum, which is 26.98 g/mol.

molar mass of Al = 26.98 g/mol

moles of Al = mass of Al / molar mass of Al
= 20.0 g / 26.98 g/mol

Step 2: Use the balanced chemical equation to determine the molar ratio between aluminum and aluminum chloride.
According to the balanced equation, 2 moles of Al react with 2 moles of AlCl3.

Step 3: Calculate the moles of aluminum chloride (AlCl3) using the molar ratio.
moles of AlCl3 = moles of Al

So, in this case, the number of moles of aluminum chloride (AlCl3) that could be produced from 20.0 g of aluminum would be the same as the moles of aluminum, which is:

moles of AlCl3 = 20.0 g / 26.98 g/mol

2) If you have excess aluminum, how many moles of aluminum chloride could be produced from 25.0 g of chlorine gas (Cl2)?

Step 1: Convert the mass of chlorine gas (Cl2) to moles.
To do this, we need to know the molar mass of chlorine gas, which is 70.90 g/mol.

molar mass of Cl2 = 70.90 g/mol

moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 25.0 g / 70.90 g/mol

Step 2: Use the balanced chemical equation to determine the molar ratio between chlorine gas and aluminum chloride.
According to the balanced equation, 3 moles of Cl2 react with 2 moles of AlCl3.

Step 3: Calculate the moles of aluminum chloride (AlCl3) using the molar ratio.
moles of AlCl3 = (3/2) * moles of Cl2

So, in this case, the number of moles of aluminum chloride (AlCl3) that could be produced from 25.0 g of chlorine gas (Cl2) would be:

moles of AlCl3 = (3/2) * (25.0 g / 70.90 g/mol)

Here is a worked example that will take care of all of your stoichiometry problems. Just follow the steps. Post your work if you get stuck.

http://www.jiskha.com/science/chemistry/stoichiometry.html

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