Question #59eb8

The reaction is #Fe_2O_3 + 3CO -> 2Fe + 3CO_2#.
And ,the molar masses(g/mol) for the substances are:
#Fe: 55.85#
#Fe_2O_3: 159.69#
#CO: 28.01#
#CO_2: 44.01#

Gas constant is
#R=8.314 (JK^-1mol^-1)#
#=0.08206 (LatmK^-1mol^-1)#

Temparature is #0℃=273.15K#.

Now we're ready to use the ideal gas law: #PV=nRT#.
The volume of gas is #V=(nRT)/P#.

[Upper row]
#1000# kg of #Fe_2O_3# is
#(1000*10^3) /159.69 = 6.262*10^3# mol.

Thus, #6.262*10^3 * 2 = 1.252*10^4# mol of #Fe# is obtained and its mass is #1.252*10^4*55.85=6.99*10^5# (g) #=699# (kg). This is right.

You need #6.262*10^3*3=1.879*10^4# mol of #CO# and its mass is #1.879*10^4*28.01=5.26*10^5# (g) #=526# (kg).

Obtained #CO_2# is also #1.879*10^4# mol and its volume is
#V=(1.879*10^4 * 0.08206 * 273.15)/1 =4.212*10^5 (L) =421.2(m^3)#.
#426.3(m^3)# in the table is a bit too large.

[Lower row]
#104.9# kg of #Fe# is #(104.9*10^3)/55.85 = 1.878*10^3# mol and to obtain this, you need #(1.878*10^3) /2 = 9.390*10^2# mol of #Fe_2O_3#. The mass of #Fe_2O_3# is #9.390*10^2*159.69=1.500 *10^5# (g) =#150.0# (kg).

The mass of #CO# is #228.9-150.0# = #78.9# (kg) and its amount of substance is #(78.9*10^3)/28.01=2.817*10^3# (mol). This is three times of that of #Fe#(#9.39*10^2# mol) and matches the chemical equation.

Obtained #CO_2# is also #2.817*10^3# mol. Its volume under #25℃,850# mmHg is
#V=(2.817*10^3*0.08206*(273.15+25))/(850/760)#
#=6.162*10^4# (L).

The "filling in the blanks" can be done with the given data and the balanced chemical equation.
#Fe_2O_3 + 3CO -> 2Fe + 3CO_2#

Converting the mass to moles for iron and the volume to moles for the #CO_2# is needed in the first line.

Iron Oxide #1000/159.7 = 6.26# kg-mol (6260 g-mol if you prefer) original

Iron #699/55.85 = 12.16# kg-mol product

The balanced reaction dictates that the #CO# required is
#3/2 xx 12.16 = 18.24# kg-mol #CO#.

This should be the same for a balanced reaction as
#3 xx 6.26 = 18.78# Kg-mol for complete reaction with the iron oxide.

The moles of #CO_2 = CO# in the equation, so we should have 18.24 kg-mol of #CO_2# product as well (from the stated Fe product amount). From the ideal gas laws we obtain:
V = nRT/P ; #V = (18240 xx 0.0802 xx 273)/1 = 399 xx 10^3L = 399m^3#

You can "mix and match" the other values to decide which ones are "incorrect" OR if you just need to fill the blank with a different number.