Question #722d3

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Answer 1 · 2016-10-05T13:41:19

#Fe_2O_3(s)+3CO(g)→2Fe(s)+3CO_2(g)#

Considering atomic masses as

#Fe->56" g/mol"#

#O->16" g/mol"#

#C->12" g/mol"#

Molar masses of reactants

#Fe_2O_3->2*56+3*16=160" g/mol"#

#CO->12+16=28" g/mol"#

As per the given equation

1 mol #Fe_2O_3# completely reacts with 3 mol #CO# or in other words

160g #Fe_2O_3# completely reacts with #3xx28=84g CO#

22.05g #Fe_2O_3# completely reacts with #=(84xx22.05)/160=11.58g CO#

But the given amount of #CO=15.56g#

So the excess amount of #CO# left after completion reaction is #(15.56-11.58)g = 3.98g#