Question

When heated KClO3 decomposes into KCl and O2 (2KClO3---->2KCl+3O2 )if this reaction produced 31.9 g of KCl,how much O2 was produced in grams

Answer 1

I think 80% sure that the answer is "(64.1 g KCl) / (74.5513 g KCl/mol) x (3 mol O2 / 2 mol KCl) x (31.99886 g O2/mol) = 41.3 g O2"
if it's wrong I am really sorry but if it's right glad to help
have a great day!!! 

Answer 2

Answer: 12 grams of oxygen

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

For $KClO_3$

Given mass = 31.9 g

Molar mass of  $KClO_3$ = 122.5 g/mol

Putting values in above equation, we get:

$\text{Moles of}KClO_3 =\frac{31.9}{122.5}=0.26moles$

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

2 moles of $KClO_3$ produces 3 moles of $O_2$

0.26 moles of $KClO_3$ produces =$\frac{3}{2}\times 0.26=0.39$ moles of $O_2$

Mass of $O_2=moles\times {\text {molar mass}}=0.39mol\times 32g/mol=12g$

Thus 12 grams of oxygen was produced.