Answer:
Maximum number of moles of AlCl3 produced is 5 moles
Explanation:
First thing's first, let's bring out the balanced chemical equation.
2Al + 3Cl2 -> 2AlCl3
Before proceeding to calculating the maximum number of moles of AlCl3 that can be formed, we have to identify the limiting reagent.
Every 2 mole of AlCl3 requires 2 moles of Al and 3 moles of Cl2.
If all of the 5 moles of Al were to be used up, there would need to be 5 × (2 / 3) or 3.333 moles of Cl2. 6 moles of Cl2 is available, this means Al is our limiting reagent.
5 mol of Al * (2 mol of Cl2 / 3 mol of Al) = 3.33 mol of Cl2
From the equation, 2 mol of Al produces 2 mol of AlCl3. This means 5 mol of Al would produce x?
2 = 2
5 = x
x = (5 * 2 ) / 2
x = 10 /2 = 5
Maximum number of moles of AlCl3 produced is 5 moles
