What is the maximum number of moles of alcl3 that can be produced from 5.0 mol al and 6.0 mol cl2? 2al + 3cl2 -> 2alcl3

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