In the formula #H_2O_2 -> H_2O + O_2#, how many grams of #O_2# are produced from the decomposition of 68 g of #H_2O_2#?

1 Answer
Jan 4, 2016

Answer:

#"mass"# (#"O"_2#)# = 32# #"g"#

Explanation:

Your equation is not correctly balanced. The correct equation is as follows:

#2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2#

First, calculate the moles of #"H"_2"O"_2# reacting. In order to do this, we must evaluate the relative molecular mass (#"M"_r#) of hydrogen peroxide:

#"M"_r# (#"H"_2"O"_2#) #= 2xx1 + 2xx16 = 34#

#"mol" = "m"/"M"_r = 68/34 = 2# #"moles"#

Next, we must compare the moles of hydrogen peroxide and oxygen gas. According to the corrected equation, this mole ratio is as shown below:

#"mol"# (#"H"_2"O"_2#) : #"mol"# (#"O"_2#)
#" "2" ":" "1#

Thus, the moles of oxygen produced in this reaction will be exactly half the moles of hydrogen peroxide reacting.

#:.# #"mol"# (#"O"_2#)# = ("mol"("H"_2"O"_2))/2 = 2/2 = 1# #"mole"#

Finally, convert from moles into mass, given that #"M"_r# (#"O"_2#)# = 32# :

#"mol" = "m"/"M"_r => "m" = "mol"xx"M"_r = 1 xx 32 = 32# #"g"#