If 15.2 g of aluminum reacts with 39.1 g of chlorine, what is the theoretical yield of AlCl?

(Hint: This will require you to identify the limiting reagent.)

To calculate the theoretical yield of AICI, we first need to determine the limiting reagent. Let's start by converting the given masses of aluminum and chlorine into moles.

Aluminum (Al):

15.2 g Al * (1 mol Al / 26.98 g Al) = 0.564 mol Al

Chlorine (Cl2):

39.1 g Cl2 * (1 mol Cl2 / 70.90 g Cl2) = 0.552 mol Cl2

Next, we need to compare the mole ratios of aluminum and chlorine to the balanced chemical equation:

2A + 3Cl2 → 2AICI

The mole ratio of aluminum to chlorine is 2:3. Since the mole ratio of aluminum to chlorine is less than 2:3, aluminum is the limiting reagent.

Now, we can use the mole ratio of aluminum to AICI to calculate the theoretical yield:

1 mol Al = 1 mol AICI

0.564 mol Al * (1 mol AICI / 2 mol Al) = 0.282 mol AICI

Finally, we can convert the moles of AICI to grams:

0.282 mol AICI * (62.5 g AICI / 1 mol AICI) = 17.6 g AICI

Learn more about calculating theoretical yield here:

https://brainly.com/question/31845156

#SPJ14

If 15.2 g of aluminum reacts with 39.1 g of chlorine, what is the theoretical yield of AlCl? (Hint: This will require you to identify the limiting reagent.)

Answer :

Final answer:

Explanation:

Other Questions

If 15.2 g of aluminum reacts with 39.1 g of chlorine, what is the theoretical yield of AlCl?

(Hint: This will require you to identify the limiting reagent.)