Question #b7deb

1 Answer
Nov 7, 2015

Answer:

#m_(Fe)=699g#

Explanation:

The reaction is:

#Fe_2O_3(s)+3CO(g)->2Fe(s)+3CO_2(g)#

Since #CO# is used in excess, therefore, the limiting reactant is #Fe_2O_3#.

#?g Fe(s) = underbrace(1.00xx10^3cancel(gFe_2O_3)xx(1cancel(molFe_2O_3))/(160 cancel(gFe_2O_3)))_(color(blue)("From g to mol"))xxunderbrace((2cancel(molFe))/(1cancel(molFe_2O_3)))_(color(blue)("Molar Ratio"))xxunderbrace((55.9gFe)/(1cancel(molFe)))_(color(blue)("From mol to g"))=color(green)(699gFe)#

I used 3 significant figures since #1.00kg# is 3 significant figures.

Explaining the dimensional analysis used above:

The first part:
#1.00xx10^3cancel(gFe_2O_3)xx(1cancel(molFe_2O_3))/(160 cancel(gFe_2O_3))#
is used to convert #gFe_2O_3# to #molFe_2O_3#

The second part:
#(2cancel(molFe))/(1cancel(molFe_2O_3))#
is the molar ration between #Fe# and #Fe_2O_3#. This is taking from the coefficient of the balanced equation.

The third part:
#(55.9gFe)/(1cancel(molFe))#
is used to convert #molFe# to #gFe#.