When 0.607 mol of Bi2O3 reacts with excess carbon, how many (a) moles of Bi and (b) grams of CO form?

The reaction involves bismuth(III) oxide ($\text{Bi}_2\text{O}_3$) reacting with carbon (C) to produce bismuth (Bi) and carbon monoxide (CO).

$$\text{Bi}_2\text{O}_3(s) + 3\text{C}(s) \rightarrow 2\text{Bi}(s) + 3\text{CO}(g)$$

From the balanced equation, 1 mole of $\text{Bi}_2\text{O}_3$ produces 2 moles of Bi.
Given moles of $\text{Bi}_2\text{O}_3 = 0.607$ mol.

$$\text{Moles of Bi} = (0.607 \text{ mol Bi}_2\text{O}_3) \times \frac{2 \text{ mol Bi}}{1 \text{ mol Bi}_2\text{O}_3}$$ $$\text{Moles of Bi} = 1.214 \text{ mol}$$

The number of moles of Bi formed is $\boxed{1.214 \text{ mol}}$.

From the balanced equation, 1 mole of $\text{Bi}_2\text{O}_3$ produces 3 moles of CO.

$$\text{Moles of CO} = (0.607 \text{ mol Bi}_2\text{O}_3) \times \frac{3 \text{ mol CO}}{1 \text{ mol Bi}_2\text{O}_3}$$ $$\text{Moles of CO} = 1.821 \text{ mol}$$

Molar mass of CO = Molar mass of C + Molar mass of O
Molar mass of C $\approx 12.01 \text{ g/mol}$
Molar mass of O $\approx 16.00 \text{ g/mol}$

$$\text{Molar mass of CO} = 12.01 \text{ g/mol} + 16.00 \text{ g/mol} = 28.01 \text{ g/mol}$$ $$\text{Mass of CO} = \text{Moles of CO} \times \text{Molar mass of CO}$$ $$\text{Mass of CO} = 1.821 \text{ mol} \times 28.01 \text{ g/mol}$$ $$\text{Mass of CO} \approx 51.006 \text{ g}$$

The mass of CO formed is $\boxed{51.01 \text{ g}}$.