Chalcocite[copper(I) sulfide] is “roasted” (heated strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. During the roasting process, how many grams of sulfur dioxide formed when 10.0 mol of copper(I) sulfide reacts?

To solve this problem, we first need to determine the stoichiometry of the reaction. From the balanced chemical equation, we can see that 2 moles of Cu2S produces 2 moles of SO2.

Therefore, 1 mole of Cu2S produces 1 mole of SO2.

Given that 10.0 mol. of Cu2S reacts,

10.0 mol. Cu2S x (1 mol. SO2 / 1 mol. Cu2S) = 10.0 mol. SO2

Now we need to convert the moles of SO2 to grams:

10.0 mol SO2 x (64.06 g / 1 mol) = 640.6 g

Therefore, 640.6 grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide reacts.