Question

9.0 mol Al produces up to 4.5 mol Al2O3 and 6.0 mol O2 produces up to 4.0 mol Al2O3. How many moles of Al2O3 form during the reaction? [?] mol Al₂O3 Give your answer to the tenths place. mol Al₂O3 Enter​

Answer 1

Step-by-step explanation:

The given information tells us that Al produces a maximum of 4.5 mol of Al2O3 and O2 produces a maximum of 4.0 mol of Al2O3. The balanced chemical equation for the reaction is:

4 Al + 3 O2 → 2 Al2O3

From the balanced equation, we see that 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3. Therefore, the ratio of moles of Al to moles of Al2O3 is 4:2 or 2:1.

If 9.0 mol of Al is available, we can calculate the maximum moles of Al2O3 that can be produced as follows:

9.0 mol Al × (2 mol Al2O3 / 4 mol Al) = 4.5 mol Al2O3

So, 4.5 mol of Al2O3 can be produced from 9.0 mol of Al.

We can also calculate the maximum moles of Al2O3 that can be produced from 6.0 mol of O2:

6.0 mol O2 × (2 mol Al2O3 / 3 mol O2) = 4.0 mol Al2O3

Since we are looking for the number of moles of Al2O3 that form during the reaction (which is less than or equal to the maximum values), the limiting reactant will determine the actual yield of Al2O3.

The limiting reactant is the one that produces the smaller amount of Al2O3, which is O2 in this case. Therefore, 4.0 mol of Al2O3 will be produced during the reaction.