Question

Question #722d3

Answer 1

`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`