Question

A chemist reacts 10 grams of copper and 3.21 grams of sulfur to make 9.56 grams of a blue powder. He also has some leftover copper. How much of each element should he react to make 100.4 grams of the blue powder with no leftovers?

Answer 1

To solve this problem, we need to determine the ratio of copper to sulfur in the blue powder formed. The blue powder likely consists of copper sulfide (Cu_xS_y), and in this case, we can assume it is primarily Cu₂S (copper(I) sulfide) for simplicity.

From the initial reaction:

1. **Initial Reaction**:
- Copper mass = 10 grams
- Sulfur mass = 3.21 grams
- Product mass = 9.56 grams

2. **Determine Moles**:
- Molar mass of copper (Cu) = 63.55 g/mol
- Molar mass of sulfur (S) = 32.07 g/mol
- Molar mass of Cu₂S = 63.55*2 + 32.07 = 158.17 g/mol

Calculate the moles of copper:
\[
\text{Moles of Cu} = \frac{10 \text{ g}}{63.55 \text{ g/mol}} \approx 0.157 moles
\]

Calculate the moles of sulfur:
\[
\text{Moles of S} = \frac{3.21 \text{ g}}{32.07 \text{ g/mol}} \approx 0.100 moles
\]

3. **Determine the Stoichiometry**:
The balanced equation for the formation of Cu₂S is:
\[
2 \text{Cu} + \text{S} \rightarrow \text{Cu}_2\text{S}
\]

From the equation, the stoichiometric ratio is 2 moles of Cu to 1 mole of S (2:1 ratio).

4. **Finding the Limiting Reagent**:
Using the moles we calculated:
- According to the 2:1 ratio, 0.157 moles of Cu would need:
\[
0.157 \text{ moles Cu} \times \frac{1 \text{ mole S}}{2 \text{ moles Cu}} = 0.0785 \text{ moles S}
\]
- We have 0.100 moles of S available, which is sufficient. Therefore, copper (Cu) is the limiting reagent.

5. **Products from Reaction**:
The maximum amount of Cu₂S generated from 0.157 moles of Cu can be calculated:
- Since 2 moles of Cu produce 1 mole of Cu₂S, 0.157 moles of Cu will produce:
\[
0.157 \text{ moles Cu} \times \frac{1 \text{ mole Cu}_2\text{S}}{2 \text{ moles Cu}} = 0.0785 \text{ moles Cu}_2\text{S}
\]

The mass of Cu₂S produced will be:
\[
\text{Mass} = 0.0785 \text{ moles} \times 158.17 \text{ g/mol} \approx 12.43 \text{ g}
\]

6. **Scale up to Desired Product Mass**:
We want to produce 100.4 grams of the blue powder (Cu₂S):
- To find the moles of Cu₂S needed:
\[
\text{Moles of Cu}_2\text{S} = \frac{100.4 \text{ g}}{158.17 \text{ g/mol}} \approx 0.634 moles
\]

Since it takes 2 moles of Cu for 1 mole of Cu₂S, we need:
\[
\text{Moles of Cu} = 0.634 \text{ moles Cu}_2\text{S} \times 2 = 1.27 \text{ moles Cu}
\]

For sulfur, using the 1:2 ratio from Cu₂S, we need:
\[
\text{Moles of S} = 0.634 \text{ moles Cu}_2\text{S} \times 1 = 0.634 \text{ moles S}
\]

7. **Convert Moles to Grams**:
Mass of copper required:
\[
1.27 \text{ moles Cu} \times 63.55 \text{ g/mol} \approx 80.80 \text{ g}
\]

Mass of sulfur required:
\[
0.634 \text{ moles S} \times 32.07 \text{ g/mol} \approx 20.30 \text{ g}
\]

### Summary:
To produce 100.4 grams of the blue powder (Cu₂S) with no leftovers, the chemist should react:
- **80.80 grams of copper (Cu)**
- **20.30 grams of sulfur (S)**