You are given 28.0 of aluminum and 33.0 of chlorine gas.

Part A

If you had excess chlorine, how many moles ofof aluminum chloride could be produced from 28.0 of aluminum?

28.0 what Al?

33.0 what Cl2?
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To determine how many moles of aluminum chloride can be produced from 28.0 grams of aluminum, we need to use the balanced chemical equation for the reaction between aluminum and chlorine to form aluminum chloride.

The balanced chemical equation is as follows:

2 Al + 3 Cl2 -> 2 AlCl3

From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride.

To find the number of moles of aluminum chloride produced, we first need to convert the mass of aluminum into moles using its molar mass. The molar mass of aluminum is 26.98 g/mol.

Mass of aluminum = 28.0 g
Molar mass of aluminum = 26.98 g/mol

Number of moles of aluminum = mass of aluminum / molar mass of aluminum
Number of moles of aluminum = 28.0 g / 26.98 g/mol

Now, we can use the stoichiometry of the balanced equation to find the number of moles of aluminum chloride produced.

2 moles of aluminum -> 2 moles of aluminum chloride

Therefore, the number of moles of aluminum chloride produced is equal to the number of moles of aluminum.

Number of moles of aluminum chloride = 28.0 g / 26.98 g/mol

So, if you had excess chlorine, you would be able to produce 28.0 moles of aluminum chloride from 28.0 grams of aluminum.

To determine the number of moles of aluminum chloride that could be produced from 28.0 g of aluminum, we need to first balance the chemical equation for the reaction between aluminum and chlorine gas. The balanced equation is:

2Al + 3Cl₂ → 2AlCl₃

From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.

Now, we can calculate the molar mass of aluminum (Al) and aluminum chloride (AlCl₃):
- The molar mass of aluminum (Al) is 26.98 g/mol.
- The molar mass of aluminum chloride (AlCl₃) is the sum of the atomic masses of aluminum (26.98 g/mol) and chlorine (35.45 g/mol). Therefore, it is 26.98 + (3 x 35.45) = 133.34 g/mol.

To find the number of moles of aluminum chloride produced from 28.0 g of aluminum, we can use the formula:

Moles = Mass / Molar mass

Moles of aluminum chloride = 28.0 g / 133.34 g/mol

Now, let's calculate the number of moles of aluminum chloride:

Moles of aluminum chloride = 0.2100 mol

Therefore, if you had excess chlorine, you could produce 0.2100 moles of aluminum chloride from 28.0 g of aluminum.