4 KO2(s) + 2 H2O(l) 4 KOH(s) + 3 O2(g)
If reaction vessel contains 0.25 mol KO2 and 0.15 mol of H2O what is limiting reactant? How
many moles of oxygen can be produced?
the equation tells you that you will need twice as many moles of KO2 as H2O.
So figure out how many moles of each you have, and see which will run out first. That will limit the reaction.
You will get either
3/4 as many moles of O2 as KO2
or
3/2 as many moles of O2 as H2O
Two separate ways to do this.
4 KO2(s) + 2 H2O(l) 4 KOH(s) + 3 O2(g)
0.25 mol..... 0.15 mol
1. short way. Choose either, let's pick KO2. Convert 0.25 mols to water needed.
0.25 mols KO2 x (2 mols H2O/4 mols KO2) = 0.125 mols H2O needed. Do you have that much H2O. Yes, therefore, KO2 is the limiting reagent (LR) and H2O is the excess reagent (ER). Or choose the other one. Convert 0.15 moles H2O to moles KO2 needed.
0.15 moles H2O x (4 mols KO2/2 moles H2O) = 0.30 moles KO2 needed. Do you have that much KO2. No, you have only 0.25 moles KO2; therefore, KO2 is the LR.
2. the long way. Pick a product II'll pick KOH) and calculate the moles possible for each reactant.
First, KO2. 0.25 mols KO2 x (4 moles KOH/4 mol KO2) = 0.25 mols KOH produced.
Next, H2O. 0.15 moles H2O x (4 mols KOH/2 mols H2O) = 0.30 mol KOH produced.
The LR is ALWAYS the smaller of the two; therefore, KO2 is the LR and H2O is the ER.
To determine the limiting reactant, we need to compare the stoichiometric ratios of the reactants.
1. Start by writing the balanced chemical equation:
4 KO2(s) + 2 H2O(l) → 4 KOH(s) + 3 O2(g)
2. Find the number of moles of each reactant:
Number of moles of KO2 = 0.25 mol
Number of moles of H2O = 0.15 mol
3. Determine the mole ratio between KO2 and H2O using the coefficients in the balanced equation:
For KO2: 4 mol KO2 : 2 mol H2O
For H2O: 2 mol H2O : 4 mol KO2
4. Convert the moles of each reactant to a common mole ratio:
Number of moles of KO2 = 0.25 mol * (2 mol H2O / 4 mol KO2) = 0.125 mol H2O
Number of moles of H2O = 0.15 mol * (4 mol KO2 / 2 mol H2O) = 0.30 mol KO2
From the calculations, we can see that the KO2 is present in excess, and the H2O is the limiting reactant because there is less H2O than what is required for the reaction to proceed completely.
5. Calculate the number of moles of oxygen produced based on the stoichiometry:
Using the mole ratio from the balanced equation, we know that for every 4 mol KO2, 3 mol O2 is produced.
Therefore, using the number of moles of H2O as the limiting reactant, we have:
0.15 mol H2O * (3 mol O2 / 2 mol H2O) = 0.225 mol O2
Thus, the limiting reactant is H2O, and 0.225 moles of oxygen gas can be produced.
To determine the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio given in the balanced chemical equation.
According to the balanced equation:
4 KO2(s) + 2 H2O(l) → 4 KOH(s) + 3 O2(g)
From the equation, we can see that the stoichiometric ratio between KO2 and O2 is 4:3, which means that for every 4 mol of KO2, we can produce 3 mol of O2.
1. Calculate the number of moles of O2 that can be produced from the given amount of KO2:
Given: 0.25 mol KO2
Using the stoichiometry, we can calculate the moles of O2:
Moles of O2 = (0.25 mol KO2) * (3 mol O2 / 4 mol KO2)
Moles of O2 = 0.1875 mol O2
2. Calculate the number of moles of O2 that can be produced from the given amount of H2O:
Given: 0.15 mol H2O
Using the stoichiometry, we can calculate the moles of O2:
Moles of O2 = (0.15 mol H2O) * (3 mol O2 / 2 mol H2O)
Moles of O2 = 0.225 mol O2
3. Compare the moles of O2 produced from KO2 and H2O:
Moles of O2 from KO2 = 0.1875 mol
Moles of O2 from H2O = 0.225 mol
Comparing the values, we can see that the moles of O2 produced from KO2 (0.1875 mol) are less than the moles of O2 produced from H2O (0.225 mol). Therefore, the limiting reactant is KO2.
So, the limiting reactant is KO2 and the maximum amount of O2 that can be produced is 0.1875 mol.