Fe3O4 (s) + 4CO (g) -> 3Fe (s) + 4CO2 (g) 694.6 g of Fe3O4 are mixed with CO (V=

Question

Fe3O4 (s) + 4CO (g) -> 3Fe (s) + 4CO2 (g)

694.6 g of Fe3O4 are mixed with CO (V= 301 L, P=1.2 atm, T = 28 C) and allowed to react. After the reaction is complete, the gases are separated and measured. What is the volume of CO2 formed if the pressure is still 1.2 atm but the temperature has risen to 35 C?

Molecular weight of Fe3O4 is 694.6 g/mole. R = 0.08206 L atm/mole K

I think this is a PV = nRT problem but I'm not sure how to deal with the change in temperature. I have all of the values for the equation, except for n, but I can calculate that. So I'm not really sure how I can use the equation to solve the problem.

Explanation / Answer

Fe3O4(s) + 4 CO (g) => 3 Fe(s) + 4 CO2(g)

Theoretical moles of Fe3O4 : CO = 1 : 4


Moles of Fe3O4 = mass/molar mass = 694.6/231.53 = 3.00 mol

Moles of CO = n = PV/RT = 1.2 x 301/(0.08206 x (273.15 + 28)) = 14.62 mol

Actual moles of Fe3O4 : CO = 3.00 : 14.62 = 1 : 4.873


Since CO is in excess => Fe3O4 is the limiting reactant

Moles of CO2 = 4 x moles of Fe3O4

= 4 x 3.00 = 12.00 mol


Ideal gas equation: PV = nRT

Volume of CO2 = V = nRT/P

= 12.00 x 0.08206 x (273.15 + 35)/1.2

= 253 L

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