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:
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Initial Reaction:
- Copper mass = 10 grams
- Sulfur mass = 3.21 grams
- Product mass = 9.56 grams
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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: \[ \text{Moles of S} = \frac{3.21 \text{ g}}{32.07 \text{ g/mol}} \approx 0.100 moles \]
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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).
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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.
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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: \[ \text{Mass} = 0.0785 \text{ moles} \times 158.17 \text{ g/mol} \approx 12.43 \text{ g} \]
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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: \[ \text{Moles of S} = 0.634 \text{ moles Cu}_2\text{S} \times 1 = 0.634 \text{ moles S} \]
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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)