Question

carbon monoxide (28.01 g/mol) is used in refining iron(iii) oxide (159.7 g/mol) to produce the metal. which species would be the limiting reactant for the production of iron if 140 g co and 8.00 × 10² g fe₂o₃ were allowed to react? fe₂o₃(s) + 3 co(g) → 2 fe(s) + 3 co₂(g) a) co₂ b) o c) fe d) fe₂o₃ e) co

Explanation:

Step1: Calculate moles of reactants

• Moles of \(CO\): \(n_{CO}=\frac{m}{M}=\frac{140\ g}{28.01\ g/mol}\approx5\ mol\)
• Moles of \(Fe_2O_3\): \(n_{Fe_2O_3}=\frac{m}{M}=\frac{800\ g}{159.7\ g/mol}\approx5\ mol\)

Step2: Use stoichiometry from the balanced equation \(Fe_2O_3(s)+3CO(g)\to2Fe(s)+3CO_2(g)\)

• From the equation, the mole ratio of \(Fe_2O_3\) to \(CO\) is \(1:3\)
• For \(5\ mol\) of \(Fe_2O_3\), the moles of \(CO\) required would be \(n = 3\times5\ mol=15\ mol\)
• But we have only \(5\ mol\) of \(CO\)
• For \(5\ mol\) of \(CO\), the moles of \(Fe_2O_3\) required would be \(n=\frac{5\ mol}{3}\approx1.67\ mol\)

Since we have less \(CO\) than what is required for the given \(Fe_2O_3\), \(CO\) is the limiting reactant.

Answer:

e) \(CO\)