Question

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Answer 1

Wait

Answer 2

KCLO3 decomposes by two parallel reaction
\[\huge(i) 2KCLO_3 \rightarrow 2KCL + 3O_2\]
\[\huge (ii) 4KCLO_3\rightarrow 3KCL0_4 +KCL\]
If 3 moles of O2 and 1 mol of KCLO4 is produced along with other products then determine initial moles of KCLO3

Answer 3

@Kainui

Answer 4

By the first reaction, 3 moles of O2 is produced from 2 moles of KCLO3
By the second reaction, 1 mole of KCLO4 is produced from 4/3 moles of KCLO3

Answer 5

Total: (2 + 4/3) moles of KCLO3

Answer 6

His answwer is correct but , i have a question

Answer 7

I am here, yes. The answer seems good. What's up?

Answer 8

In the second reaction why did we not consider the molar co-efficients , @aum is online let's ask him

Answer 9

No, he did consider them.

See in the second reaction it says 4 mols of KClO3 makes 3 mols KClO4

So this means we can set these equal to each other: \[\LARGE 4 \ mols \ of \ KClO_3 = 3\ mols \ KClO_4\]So now we can create a conversion factor out of this\[\LARGE \frac{4 \ mols \ of \ KClO_3}{ 3\ mols \ KClO_4}=1\] since we're given 1 mol of KClO4 we can multiply it by this form of 1 to get it in terms of what we want\[\LARGE \frac{4 \ mols \ of \ KClO_3}{ 3\ mols \ KClO_4}*1 \ mol \ KClO_4\]\[\LARGE =\frac{4}{3}\ mols \ of \ KClO_3\]

It's all in the units again haha.

Answer 10

O_o i see

Answer 11

In fact this method is very general and works for many chemistry and physics problems where some type of conversion needs to take place. You can literally divide out the units as if they're variables or numbers and is incredibly useful.

A good example I like to show people is:
\[\LARGE \frac{elephants}{elephants}=1\]

Answer 12

thanks all