Consider the chemical reaction, 2Al + 3Cl2 → 2AlCl3. Here, 56 g of Al can produce 415.14 g of AlCl3 (Molar masses: Al = 26.98 g; Cl = 35.45 g).

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To determine whether the statement is true or false, we need to calculate the theoretical yield of AlCl3 using the given molar masses and stoichiometry of the reaction.

First, we need to convert the mass of Al to moles using its molar mass:

Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 56 g / 26.98 g/mol
Moles of Al = 2.075 mol

According to the balanced equation, the ratio of moles of Al to moles of AlCl3 is 2:2. This means that for every 2 moles of Al, we should obtain 2 moles of AlCl3.

Now, we can calculate the moles of AlCl3 using the moles of Al:

Moles of AlCl3 = Moles of Al
Moles of AlCl3 = 2.075 mol

Finally, we can convert the moles of AlCl3 to grams using its molar mass:

Mass of AlCl3 = Moles of AlCl3 × Molar mass of AlCl3
Mass of AlCl3 = 2.075 mol × (26.98 g + (3 × 35.45 g))
Mass of AlCl3 = 2.075 mol × 133.78 g/mol
Mass of AlCl3 = 277.45 g

Now, we can compare the calculated theoretical yield of AlCl3 to the given value of 415.14 g.

The statement is false because the calculated theoretical yield of AlCl3 (277.45 g) is less than the given value of 415.14 g.