How many grams of AlCl3 are produced when 53.0 g of Al reacts?
To determine the number of grams of AlCl3 produced when the given amount of Al reacts, we need to balance the chemical equation for the reaction between Al and AlCl3.
The balanced chemical equation for the reaction is:
2 Al + 3 Cl2 → 2 AlCl3
From the balanced equation, we can see that 2 moles of Al react to produce 2 moles of AlCl3. Therefore, the molar ratio between Al and AlCl3 is 2:2 or 1:1.
Next, we need to calculate the moles of Al present in the given amount of Al.
Molar mass of Al = 26.98 g/mol
Moles of Al = (Given mass of Al) / (Molar mass of Al)
= 53.0 g / 26.98 g/mol
= 1.9642 mol (rounded to four decimal places)
Since the molar ratio between Al and AlCl3 is 1:1, the number of moles of AlCl3 produced is also 1.9642 mol.
Finally, we can calculate the mass of AlCl3 produced:
Molar mass of AlCl3 = (1 x molar mass of Al) + (3 x molar mass of Cl)
= (1 x 26.98 g/mol) + (3 x 35.45 g/mol)
= 26.98 g/mol + 106.35 g/mol
= 133.33 g/mol
Mass of AlCl3 = (Number of moles of AlCl3) x (Molar mass of AlCl3)
= 1.9642 mol x 133.33 g/mol
= 261.97 g
Therefore, when 53.0 g of Al reacts, 261.97 g of AlCl3 are produced.
To determine the number of grams of AlCl3 produced when 53.0 g of Al reacts, we need to use the concept of stoichiometry.
The balanced chemical equation for the reaction between Al and Cl2 to form AlCl3 is:
2Al + 3Cl2 → 2AlCl3
From the balanced equation, we can see that 2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3.
To solve the problem, follow these steps:
Step 1: Convert the given mass of Al (53.0 g) into moles.
To do this, you need to know the molar mass of Al, which is approximately 26.98 g/mol.
Number of moles of Al = Mass of Al / Molar mass of Al
= 53.0 g / 26.98 g/mol
≈ 1.967 moles
Step 2: Use the mole ratio from the balanced equation to find the number of moles of AlCl3 produced.
From the balanced equation, we can see that 2 moles of Al reacts with 2 moles of AlCl3.
Number of moles of AlCl3 = Number of moles of Al × (2 moles AlCl3 / 2 moles Al)
= 1.967 moles × (2 moles AlCl3 / 2 moles Al)
= 1.967 moles
Step 3: Convert the number of moles of AlCl3 into grams.
To do this, you need to know the molar mass of AlCl3, which is approximately 133.34 g/mol.
Mass of AlCl3 = Number of moles of AlCl3 × Molar mass of AlCl3
= 1.967 moles × 133.34 g/mol
≈ 261.96 g
Therefore, approximately 261.96 grams of AlCl3 are produced when 53.0 grams of Al reacts.
moles of Al=53/27
moles of AlCl3 = same
mass AlCl3=molesAlCl3*molmassAlCl3