(a)(i) Volume of CO (g) = 16.2 L
Pressure of CO (g) = 1.50 atm
Temperature of CO (g) = 200.ºC
Convert the temperature to the absolute scale.
T = (273 + 200.)K = 473 K
Use the ideal gas to determine the number of moles of CO.
PV = nRT
where the symbols have their usual meanings.
Therefore,
n = PV/RT
Plug in values and get
n = (1.50 atm)*(16.2 L)/(0.082 L-atm/mol.K)(473 K)
= 0.6265 mol
≈ 0.626 mol (ans, correct to 3 sig, figs).
(ii) The balanced chemical reaction is given as
Fe2O3 (s) + 3 CO (g) ----------> 2 Fe (s) + 3 CO2 (g)
As per the stoichiometric equation,
1 mole Fe2O3 = 3 moles CO.
Mass of Fe2O3 taken = 15.39 g.
The atomic masses are
Fe: 55.845 g/mol
O: 15.999 g/mol
Gram molar mass of Fe2O3 = (2*55.845 + 3*15.999) g/mol
= 159.687 g/mol
Mol(s) Fe2O3 corresponding to 15.39 g = (15.39 g)/(159.687 g/mol)
= 0.09637 mol
≈ 0.0964 mol (correct to 3 sig. figs)
Determine the limiting reactant.
Fe2O3: (0.0964 mol Fe2O3)*(3 mols CO)/(1 mol Fe2O3) = 0.2892 mol CO
CO: (0.626 mol)*(1 mol Fe2O3)/(3 mols CO) = 0.20866 mol Fe2O3 ≈ 0.2087 mol Fe2O3.
We have only 0.0964 mol Fe2O3 and more than 0.2892 mol CO. Therefore, Fe2O3 is the limiting reactant.
(iii) As per the balanced stoichiometric equation,
1 mol Fe2O3 = 2 mols Fe.
Therefore,
Mols Fe corresponding to 0.0964 mol Fe2O3 = (0.0964 mol Fe2O3)*(2 mols Fe)/(1 mol Fe2O3)
= 0.1928 mol Fe.
≈ 0.193 mol Fe (correct to 3 sig, figs)
The moles of Fe formed = 0.193 mol (ans).
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