Final answer:
The statement is true; cobalt metal can indeed be prepared by reducing cobalt(II) oxide with carbon monoxide. The balanced chemical reaction is CoO(s) + CO(g) = Co(s) + CO₂(g), and at equilibrium with a given [CO₂] and Kc value, the concentration of CO can be calculated.
Step-by-step explanation:
The statement that cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide is indeed true. The reduction process is a chemical reaction where cobalt(II) oxide (CoO) reacts with carbon monoxide (CO) to produce cobalt metal (Co) and carbon dioxide (CO₂). The balanced chemical equation for this reduction process is:
CoO(s) + CO(g) = Co(s) + CO₂(g)
This type of reaction is a redox process, where carbon monoxide serves as the reducing agent that accepts oxygen from the cobalt oxide, turning into carbon dioxide, while the cobalt ions gain electrons to become cobalt metal. This approach is one of the classical methods for extracting metals from their ores, relying on the reduction in mass due to the loss of oxygen as metals are reduced from their oxide forms.
In terms of calculating the equilibrium concentrations of reactants based on a given equilibrium constant (Kc) and the concentration of products, we use the information provided:
Kc = 4.90 × 10² at 550 °C
For the reaction at equilibrium, with [CO₂] = 0.100 M, and since solid cobalt(II) oxide does not affect equilibrium concentrations, we can express the equilibrium constant as:
Kc = [CO₂] / [CO]
By rearranging and solving for [CO], we find:
[CO] = [CO₂] / Kc = 0.100 M / 4.90 × 10² = 2.04 × 10⁻⁴ M
Thus, the concentration of CO remaining in the equilibrium mixture is approximately 2.04 × 10⁻⁴ M.