Tori71988
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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

Respuesta :

halaalnajar
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: 12 grams of oxygen

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

For [tex]KClO_3[/tex]

Given mass = 31.9 g

Molar mass of  [tex]KClO_3[/tex] = 122.5 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of}KClO_3 =\frac{31.9}{122.5}=0.26moles[/tex]

[tex]2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)[/tex]

2 moles of [tex]KClO_3[/tex] produces 3 moles of [tex]O_2[/tex]

0.26 moles of [tex]KClO_3[/tex] produces =[tex]\frac{3}{2}\times 0.26=0.39[/tex] moles of [tex]O_2[/tex]

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

Thus 12 grams of oxygen was produced.

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