2Al+3Cl2=2AlCl3

calculate the amount(number) of excess reagen units remaining when 4.03x10^23 Al atoms and 6.42x10^23 Cl2 molecules react?

please help!!!!! thanks

For each Al, one needs 1.5 Cl2

So, one needs 1.5*4.03E23 or 6.045E23 Cl2.
Thus, there is an excess of Cl2.

To determine the number of excess reagent units remaining, we need to first determine the limiting reagent in the reaction. The limiting reagent is the reactant that is consumed completely and determines the maximum amount of product that can be formed.

Let's start by calculating the number of moles for each reactant using the given quantities:

Number of moles of Al:
4.03 x 10^23 Al atoms x (1 mole/6.022 x 10^23 atoms) = 0.67 moles of Al

Number of moles of Cl2:
6.42 x 10^23 Cl2 molecules x (1 mole/Avogadro's number) = 0.11 moles of Cl2

Next, we need to determine the stoichiometric ratio between Al and Cl2 in the balanced equation. From the balanced equation: 2Al + 3Cl2 → 2AlCl3, we can see that the ratio of Al to Cl2 is 2:3.

Comparing the moles of Al and Cl2, we see that there is a 2:3 ratio. However, the actual ratio we have is lower, with 0.67 moles of Al to 0.11 moles of Cl2.

From the above calculation, we can conclude that Cl2 is the limiting reagent because there is less of it compared to what is needed for the reaction.

Now, let's calculate the amount of excess reagent (Al) remaining after the reaction.

To determine the number of moles of Cl2 required to react completely with 0.67 moles of Al, we will use the stoichiometric ratio of 2:3 (Al:Cl2) from the balanced equation.

(0.67 moles Al) x (3 moles Cl2/2 moles Al) = 1.01 moles Cl2 (required)

From the given, we have 0.11 moles of Cl2, which is less than the amount required (1.01 moles). This means that all the Cl2 will be consumed in the reaction.

To calculate the amount of excess reagent (Al) remaining, we need to subtract the amount of Al that reacted from the initial amount given.

Initial moles of Al = 0.67 moles (as calculated before)

Amount of Al that reacted = 0.67 moles - 0.67 moles (which reacted with Cl2) = 0 moles

Therefore, there are 0 moles of excess Al remaining after the reaction.

If you want to convert the moles of Al to the number of Al atoms remaining, you can use Avogadro's number (6.022 x 10^23 atoms/mole).

Number of remaining Al atoms = 0 moles x (6.022 x 10^23 atoms/mole) = 0 atoms

Therefore, when 4.03x10^23 Al atoms and 6.42x10^23 Cl2 molecules react, there are 0 moles of excess Al remaining, which can be converted to 0 atoms.

Hope this explanation helps! Let me know if you have any further questions.