Question

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

Answer 1

`m_(O2)=192g O_2`

Step-by-step 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* (3 mol O2)/(2 mol KClO3)*(32 g)/(mol O2)`

`m_(O2)=192g O_2`

Answer 2

from the equation above the reacting ratio of KClO3 to O2 is 2:3 therefore the number of moles of oxygen produced is ( 4 x3)/2 = 6 moles since four moles of KClO3 was consumed
mass=relative formula mass x number of moles
That is 32g/mol x 6 moles =192grams