Pure iron can be produced by the reaction between magnetite (Fe3O4) and hydrogen. Fe3O4(s) + 4H2(g) 🡺 3Fe(s) + 4H2O(l)

Question

If 100.0 g of both magnetite and hydrogen are initially present:

a. Determine the limiting reagent.

b. Calculate the theoretical yield of iron

c. If only 32.0 g Fe obtained in the process, calculate the percent yield.

d. Calculate mass of the reactant in excess that remained after the process was completed.

Answer 1

Fe₃O₄(s) + 4H₂(g) ==> 3Fe(s) + 4H₂O(l) ... balanced equation

molar mass Fe₃O₄ = 231.5 g/mol

molar mass H₂ = 2.02 g/mol

atomic mass Fe = 55.9 g/mol

a). An easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value is less is the limiting reactant.

For Fe₃O₄: 100.0 g x 1 mol / 231.5 g = 0.4320 mols Fe₃O₄ (÷1->0.432)

For H₂: 100 g x 1 mol / 2.02 g = 49.50 mols H2 (÷4->12)

Clearly, Fe₃O₄ is in limiting supply

b). Theoretical yield of Fe is determined by the moles of the limiting reactant.

0.4320 mols Fe₃O₄ x 3 mols Fe / mol Fe₃O₄ x 55.9 g Fe / mol Fe = 72.45 g Fe = theoretical yield

c). Percent yield

%yield = actual yield / theoretical yield (x100%)

%yield = 32.0 g / 72.45 g (x100%) = 44.2% yield

d). Excess reactant is H₂, and mass of H₂ remaining is....

initial mass H₂ - mass H₂ used = mass H₂ remaining

initial mass H₂ = 100.0 g

mass H₂ used = 0.4320 mols Fe₃O₄ x 4 mols H₂ / mol Fe₃O₄ x 2.02 g H2/mol = 3.49 g H₂ used

mass H₂ remaining = 100.0 g - 3.49 g = 96.5 g H₂ remaining