Question

Copper is obtained from suflide ores such as chalcocite (copper(i) sulfide) through the a multistep process. after the ore is ground, it is heated strongly with oxygen gas, roasting it, forming powdered copper(i) oxide and gaseous sulfur dioxide.

Cu₂S(s) + O₂(g) → Cu₂O(s) + SO₂(g) (unbalanced)
how many grams of sulfur dioxide form when 10.0 mol of copper(i) sulfide reacts?

Answer 1

When 10.0 moles of copper(I) sulfide react in roasting, 640.7 grams of sulfur dioxide are formed, based on the balanced equation 2 Cu2S + 3 O2 → 2 Cu2O + 2 SO2 and the molar mass of SO2.

Step-by-step explanation:

The question asks how many grams of sulfur dioxide form when 10.0 mol of copper(I) sulfide reacts in the process of roasting, which involves heating ore with oxygen to form copper(I) oxide and sulfur dioxide. The chemical reaction is given as:

Cu2S(s) + O2(g) → Cu2O(s) + SO2(g)

First, we need to balance the chemical equation:

2 Cu2S(s) + 3 O2(g) → 2 Cu2O(s) + 2 SO2(g)

From the balanced equation, we see that 2 moles of Cu2S produce 2 moles of SO2. Therefore, 10.0 moles of Cu2S will produce 10.0 moles of SO2 since the ratio is 1:1.

Next, we convert moles of SO2 to grams. The molar mass of SO2 is approximately 64.07 g/mol (32.07 for sulfur and 16.00 for each of the two oxygens).

(Moles of SO2) × (Molar mass of SO2) = Mass of SO2 in grams

10.0 mol × 64.07 g/mol = 640.7 grams

Therefore, 10.0 moles of copper(I) sulfide reacting will form 640.7 grams of sulfur dioxide.