Chemistry: The following equation is one way to prepare oxygen in a lab. 2kclo3 → 2kcl + 3o2 molar mass info: mm o2 = 32 g/mol mm kcl = 74.55 g/mol mm kclo3 = 122.55 g/mol if 4.00 moles of kclo3 are totally consumed, how many grams of oxygen gas would be produced? 192 g 6.00 g 85.3 g 735 g

The following equation is one way to prepare oxygen

tymiahill7244

15.03.2022

$m_{O2}=192g O_2$

Explanation:

Following the reaction, for each 2 moles of KClO3 consumed completely, 3 moles of oxygen are generated.

So, if 4 moles are consumed:

$m_{O2}=4 mol KClO3* \frac{3 mol O2}{2 mol KClO3}*\frac{32 g}{mol O2}$

$m_{O2}=192g O_2$

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lilblackbird4

14.07.2022

Explanation:

They have all 8 electrons or whatever that are required to make them satisfied

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