consider the chemical reaction, 2Al + 3Cl2 -> 2AlCl3. here, 212.7 g of chlorine can produce a maximum of 266.66 g of AlCl3 (molar mass of Al = 26.98 g and Cl = 35.45 g)
true or false
Work it out.
mols Cl2 = grams/molar mass = ?
mols = ?mols Cl2 x (2 mols AlCl3/3 mols Cl2) = ? mols Cl2 x 2/3 = ?
g AlCl3 = mols AlCl3 x molar mass AlCl3.
my opinion? I think so.
True.
To determine the maximum amount of AlCl3 that can be produced, we need to calculate the stoichiometric ratio between Cl2 and AlCl3.
According to the balanced chemical equation, 2 moles of Al reacts with 3 moles of Cl2 to produce 2 moles of AlCl3. Therefore, the ratio is 3:2 between Cl2 and AlCl3.
Now, let's calculate the moles of Cl2 present in 212.7 g of chlorine:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 212.7 g / 35.45 g/mol
moles of Cl2 = 6 moles
Using the stoichiometric ratio, we can determine the moles of AlCl3 that can be produced:
moles of AlCl3 = (moles of Cl2 / 3) * 2
moles of AlCl3 = (6 moles / 3) * 2
moles of AlCl3 = 4 moles
Finally, we can convert moles of AlCl3 to grams:
mass of AlCl3 = moles of AlCl3 * molar mass of AlCl3
mass of AlCl3 = 4 moles * (26.98 g/mol + 3 * 35.45 g/mol)
mass of AlCl3 = 4 moles * 133.29 g/mol
mass of AlCl3 = 533.16 g
Therefore, the maximum amount of AlCl3 that can be produced from 212.7 g of Cl2 is 533.16 g. Since 266.66 g is less than the calculated amount, the statement is true.
To determine whether the statement is true or false, we need to calculate the maximum amount of AlCl3 that can be produced using 212.7 g of chlorine.
First, we need to convert the mass of Cl2 into moles. The molar mass of Cl2 is 35.45 g/mol * 2 = 70.9 g/mol. So, the number of moles of Cl2 is 212.7 g / 70.9 g/mol = 2.997 mol.
Next, we can use the balanced chemical equation to determine the stoichiometry of the reaction. From the equation 2Al + 3Cl2 -> 2AlCl3, we can see that it requires 3 moles of Cl2 to produce 2 moles of AlCl3.
Using a ratio, we can find the maximum amount of AlCl3 produced. Since we have 2.997 mol of Cl2, we can multiply it by the ratio 2 mol AlCl3 / 3 mol Cl2.
(2.997 mol Cl2) * (2 mol AlCl3 / 3 mol Cl2) = 1.998 mol AlCl3
Now, we need to convert the moles of AlCl3 into grams. The molar mass of AlCl3 is 26.98 g/mol (Al) + (35.45 g/mol * 3) (Cl) = 133.34 g/mol. Therefore, the mass of 1.998 mol of AlCl3 is 1.998 mol * 133.34 g/mol = 266.44 g.
Comparing this calculated value (266.44 g) to the given value (266.66 g) in the statement, we can see that they are very close. However, because of rounding in our calculations, they may not be exactly the same.
Hence, the statement "212.7 g of chlorine can produce a maximum of 266.66 g of AlCl3" can be considered as true.