Question

Potassium superoxide, ko2 is used in rebreathing gas masks to generate oxygen. 4ko2(s) + 2h2o(l) → 4koh(s) + 3o2(g) if a reaction vessel contains 0.15 mol ko2 and 0.10 mol h2o, what is the limiting reactant? How many moles of oxygen can be produced?

Answer 1

Answer : The limiting reactant is $KO_2$.

The number of moles of oxygen gas produced can be 0.1125 moles.

Solution : Given,

Moles of $KO_2$ = 0.15 mole

Moles of $H_2O$ = 0.10 mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$4KO_2(s)+2H_2O(g)\rightarrow 4KOH(s)+3O_2(g)$

From the balanced reaction we conclude that

As, 4 mole of $KO_2$ react with 2 mole of $H_2O$

So, 0.15 moles of $KO_2$ react with $\frac{0.15}{4}\times 2=0.075$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $KO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $O_2$

From the reaction, we conclude that

As, 4 mole of $KO_2$ react to give 3 mole of $O_2$

So, 0.15 moles of $KO_2$ react to give $\frac{0.15}{4}\times 3=0.1125$ moles of $O_2$

Therefore, the number of moles of oxygen gas produced can be 0.1125 moles.

Answer 2

Answer:

KO2 is the limiting reactant

0.1875 moles of oxygen can be produced

Explanation:

To begin with, we need to find the limiting reactant in the reaction, Then use it to find the number of moles of oxygen. So whether the question asked for limiting reactant or not, we should find it. Here, its easier than you think. Look at the reactants only and leave the products away for a bit. 4KO2 + 2H2O. If we divide by 2, then we are left with 2KO2, So KO is the limiting reactant.

Then after we found our limiting reactant, We just use it in a very simple way.

If 4 Moles of KO2 ⇒ 3 Moles of O2

Then 0.25 Moles of KO2⇒X Moles of O2

You should know your math but if you don't, we cross multiply as below:

0.25*3=4X

X=0.1875