4al(s)+3o2(g)---->2o3(g) limiting reactant
7.4 mol Al
or
6.5 mol O2
Your equation can't be correct.
It's oksy i got the answer 7.4 mol Al Thank you,
Maybe you can help me with the next question. :-)
To determine the limiting reactant, you need to compare the amount of each reactant to the stoichiometric coefficients (coefficients in front of the reactant formula) in the balanced equation.
Let's start by writing the balanced equation:
4Al(s) + 3O2(g) → 2O3(g)
Now, we can calculate the number of moles of products that can be formed from the given amount of each reactant.
For 7.4 mol Al:
Using the balanced equation, we see that 4 moles of Al react with 3 moles of O2 to produce 2 moles of O3.
So, we can set up a ratio:
4 mol Al / 3 mol O2 = 2 mol O3
Now, let's calculate the number of moles of O3 that can be formed from 7.4 mol Al:
2 mol O3 * (7.4 mol Al / 4 mol Al) = 3.7 mol O3
For 6.5 mol O2:
Using the balanced equation, we see that 3 moles of O2 react with 4 moles of Al to produce 2 moles of O3.
So, we can set up a ratio:
3 mol O2 / 4 mol Al = 2 mol O3
Now, let's calculate the number of moles of O3 that can be formed from 6.5 mol O2:
2 mol O3 * (6.5 mol O2 / 3 mol O2) = 4.33 mol O3
Comparing the two calculated amounts, we find:
- 3.7 mol O3 can be formed from 7.4 mol Al
- 4.33 mol O3 can be formed from 6.5 mol O2
Since we cannot have a fractional amount of a reactant, we need to round down to the nearest whole number. In this case, we need to round down 4.33 to 4. Therefore, the limiting reactant is the one that produces the lesser amount of O3, which is 6.5 mol O2.
So, the limiting reactant in this reaction is 6.5 mol O2.