if you had excess chlorine, how many moles of of aluminum chloride could be produced from 17.0g of aluminum?
Express your answer numerically in moles.
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To determine the moles of aluminum chloride that can be produced from 17.0g of aluminum, we need to use the balanced chemical equation and the molar mass of aluminum.
The balanced equation for the reaction between aluminum and chlorine to produce aluminum chloride is:
2 Al + 3 Cl2 -> 2 AlCl3
From the equation, we can see that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.
First, we need to find the moles of aluminum in 17.0g of aluminum using its molar mass. The molar mass of aluminum (Al) is approximately 26.98 g/mol.
Moles of aluminum = mass of aluminum / molar mass of aluminum
Moles of aluminum = 17.0g / 26.98 g/mol
Next, we can use the mole ratio from the balanced equation to find the moles of aluminum chloride produced. The mole ratio shows that 2 moles of aluminum produce 2 moles of aluminum chloride.
Moles of aluminum chloride = Moles of aluminum
Hence, the moles of aluminum chloride that can be produced from 17.0g of aluminum is equal to the moles of aluminum, which is approximately:
17.0g / 26.98 g/mol = 0.630 moles of aluminum chloride
To determine how many moles of aluminum chloride could be produced from 17.0g of aluminum, we need to use the concept of stoichiometry.
First, let's write a balanced chemical equation for the reaction between aluminum (Al) and chlorine (Cl2) to form aluminum chloride (AlCl3):
2 Al + 3 Cl2 -> 2 AlCl3
From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.
Next, we'll use the molar mass of aluminum to convert the given mass of aluminum (17.0g) to moles. The molar mass of aluminum (Al) is 26.98 g/mol.
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
Moles of aluminum = 17.0g / 26.98 g/mol
Moles of aluminum = 0.6303 mol (rounded to four decimal places)
Using the stoichiometry of the balanced equation, we know that 2 moles of aluminum react with 2 moles of aluminum chloride. Therefore, the number of moles of aluminum chloride produced is the same as the number of moles of aluminum:
Moles of aluminum chloride = 0.6303 mol (rounded to four decimal places)
Therefore, if you had excess chlorine, you could produce 0.6303 moles of aluminum chloride from 17.0g of aluminum.
This is the same as the NH3 problem but backwards.
Write and balance the equation.
Convert 17.0 g Al to mols. mols = grams/atomic mass.
Using the coefficients in the balanced equation, convert mols Al to mols AlCl3.