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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

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kenmyna
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

Giangiglia

Answer:

[tex]m_{O2}=192g O_2[/tex]

Explanation:

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

So, if 4 moles are consumed:

[tex]m_{O2}=4 mol KClO3* \frac{3 mol O2}{2 mol KClO3}*\frac{32 g}{mol O2}[/tex]

[tex]m_{O2}=192g O_2[/tex]

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