Question

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?

Answer 1

[AI-generated answer — Bot GPT 3.5]

The balanced chemical equation for the reaction is:

2 Cu2S + 3 O2 -> 2 Cu2O + 2 SO2

From the equation, we can see that 2 moles of copper sulfide react to produce 2 moles of sulfur dioxide.

Therefore, when 10.0 mol of copper(I) sulfide reacts, 10.0 mol / 2 = 5.0 mol of sulfur dioxide will be formed.

Now, we can calculate the mass of sulfur dioxide formed using the molar mass of SO2:

Molar mass of SO2 = 32.07 g/mol + 2 * 15.999 g/mol = 64.07 g/mol

Mass of 5.0 mol of SO2 = 5.0 mol * 64.07 g/mol = 320.35 g

Therefore, when 10.0 mol of copper(I) sulfide reacts, 320.35 grams of sulfur dioxide will be formed.