Determine the mass aluminum chloride , alcl3 produced when 112.3 g of chloride gas, cl2 is used
It's chlorINE gas is used. Won't it depend upon how much Al is used, also? I assume the author of the problem SHOULD HAVE SAID and excess of Al.Watch the caps; alcl3 doesn't mean anything. For example, Co is cobalt, CO is carbon monoxide, co is company.
2Al + 3Cl2 ==> 2AlCl3
mols Cl2 = grams/molar mass = 112.3/about 71 = ?
Using the coefficients in the balanced equation, convert mols Cl2 to mols AlCl3.
Then grams AlCl3 = mols AlCl3 x molar mass AlCl3.
To determine the mass of aluminum chloride (AlCl3) produced, we first need to know the balanced equation for the reaction between aluminum and chlorine gas:
2Al + 3Cl2 -> 2AlCl3
From the balanced equation, we can see that 2 moles of aluminum (Al) react with 3 moles of chlorine gas (Cl2) to produce 2 moles of aluminum chloride (AlCl3).
Now, let's calculate the number of moles of chlorine gas (Cl2) that you have:
First, find the molar mass of chlorine gas (Cl2), which is the sum of the molar masses of its individual atoms. The molar mass of chlorine (Cl) is approximately 35.45 g/mol, so the molar mass of Cl2 is 2(35.45) = 70.90 g/mol.
Next, calculate the number of moles of Cl2 by dividing its mass (112.3 g) by its molar mass (70.90 g/mol):
moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 112.3 g / 70.90 g/mol
≈ 1.582 mol
Now, referring back to the balanced equation, we know that 3 moles of Cl2 react to produce 2 moles of AlCl3. Therefore, the number of moles of AlCl3 produced is calculated using the mole ratio:
moles of AlCl3 = (moles of Cl2) * (2 moles AlCl3 / 3 moles Cl2)
= 1.582 mol * (2/3)
≈ 1.055 mol
Finally, we need to calculate the mass of the aluminum chloride (AlCl3) produced. The molar mass of AlCl3 is the sum of the molar masses of aluminum (Al) and chlorine (Cl). The molar mass of aluminum is approximately 26.98 g/mol, and the molar mass of chlorine is approximately 35.45 g/mol. Therefore, the molar mass of AlCl3 is (26.98 + 3(35.45)) = 133.34 g/mol.
mass of AlCl3 = (moles of AlCl3) * (molar mass of AlCl3)
= 1.055 mol * 133.34 g/mol
≈ 140.85 g
Therefore, the mass of aluminum chloride (AlCl3) produced when 112.3 g of chlorine gas (Cl2) is used is approximately 140.85 grams.