Ferric oxide may be reduced to pure iron either with carbon monoxide or with coke (pure carbon). Suppose that 150.0 lb of ferric oxide are available. How many pounds of carbon monoxide would be required to reduce the oxide?

Question

How many pounds of coke would be needed? In each case, how many pounds of pure iron would be produced?

Peyton Adams · Accepted Answer

#C(s) + 1/2O_2(g) rarr CO(g)#
#Fe_2O_3(s) + 3CO(g) rarr 2Fe(s) + 3CO_2(g)# The answer relates to the stoichiometric equation, which tells us that #159.7*g# #Fe_2O_3# reacts with #84.0*g# #CO# to give #112*g# #Fe#. You have #150*"lbs"xx454*g*"lb"^-1# #=# #68,100*g# #Fe_2O_3#. Given the equations you should be able to tells us the quantities of steel produced, and the quantity of gas required. If you have difficulties, state them here, and someone will help you.

Avery Adams · Answer

undefined To calculate the amount of carbon monoxide required to reduce 150.0 lb of ferric oxide, we need to use the stoichiometry of the reaction. The balanced chemical equation for the reduction of ferric oxide (Fe2O3) to iron (Fe) using carbon monoxide (CO) is:

3Fe2O3 + 3CO -> 2Fe3O4 + 3CO2

From the balanced equation, we see that 3 moles of CO are required to react with 3 moles of Fe2O3. The molar mass of Fe2O3 is approximately 159.69 g/mol, and the molar mass of CO is approximately 28.01 g/mol.

First, we need to convert the mass of ferric oxide to moles:

150.0 lb * (453.59 g/lb) / (159.69 g/mol) = 425.09 mol Fe2O3

Now, we can use the stoichiometry of the reaction to find the moles of CO required:

1 mole of Fe2O3 requires 3 moles of CO

So, 425.09 mol Fe2O3 * (3 mol CO / 3 mol Fe2O3) = 425.09 mol CO

Finally, we convert moles of CO to pounds:

425.09 mol CO * (28.01 g/mol) / (453.59 g/lb) = 26.26 lb CO