Question

How many grams of oxygen are formed when 8.20 moles of KOH are formed? 4 KO(s) + 2 H2O(l) → 4 KOH(s) + O2(g)

Answer 1

49.7g02* that's the answer

Answer 2

Answer : The mass of oxygen formed are 65.6 grams.

Explanation :

First we have to calculate the moles of $O_2$

The given balance chemical reaction is:

$4KO(s)+2H_2O(l)rightarrow 4KOH(s)+O_2(g)$

From the balanced chemical reaction we conclude that,

As, when 4 moles of KOH formed then 1 mole of $O_2$ will be formed

So, when 8.20 moles of KOH formed then $\frac{8.20}{4}=2.05$ mole of $O_2$ will be formed

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

Molar mass of $O_2$ = 32 g/mole

$\text{ Mass of }O_2=(2.05moles)\times (32g/mole)=65.6g$

Therefore, the mass of oxygen formed are 65.6 grams.