How many grams of KClO3 are needed to produce 50g of O2?

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SUMMARY

The discussion centers on calculating the grams of potassium chlorate (KClO3) required to produce 50 grams of oxygen (O2) through the decomposition reaction 2KClO3 → 2KCl + 3O2. The correct molar mass of KClO3 is 122.55 g/mol, leading to the conclusion that 127 grams of KClO3 is needed to yield 50 grams of O2. Participants clarified misconceptions regarding the molar mass of O2 and the proper application of stoichiometric ratios in the calculation.

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Reema
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We had this question in the lecture .. but I couldn'nt copy the steps I was trying to understand but I dind'nt looool.

How many grams of KClO3 are required to produce 50 g of 02 from the following decomposition reaction?

2KClO3 ---> 2kClO3 + 3O2


the final answer is 127


I tried to work it out by myself by making a ratio


molar mass of 2KClO3 = 148.8 and for 3O2 = 48

grams of 02 = 3*50 = 150 grams
grams of of 2KClO3 = x


x\148 : 150\48

x=265


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You are doing strange things.

Reema said:
2KClO3 ---> 2kClO3 + 3O2

Huh? I guess it should be KCl on the right.


molar mass of 2KClO3 = 148.8 and for 3O2 = 48

What do you mean by "molar mass of 3O2"? No such animal. There is a molar mass of O2. It is not 48.

grams of 02 = 3*50 = 150 grams

No, 50 grams of O2. This is a number given in the question.

I am not even going to try to understand your ratio, as numbers you put in it are already off.

Try these pages:

stoichiometric calculations in general

stoichiometric calculations using ratios
 

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