Question #967d1

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Answer 1 · 2015-03-13T01:38:48

155.5g will be formed.

#4Fe+3O_2rarr2Fe_2O_3#

So 3 mol #O_2# #rarr# 2 mol #Fe_2O_3#

Using approximate #A_r# values:

#3xx(2xx16)#g #O_2# #rarr2[(56xx2)+(3xx16)]# g #Fe_2O_3#

So:

#64grarr320g#

So:

#1grarr320/64g#

So:

#31.1grarr(320)/(64)xx31.1=155.5g#