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Calculate the volume of sulfur dioxide gas produced when 14.29 grams of [tex]\( \text{Cu}_2\text{S} \)[/tex] react with excess oxygen at [tex]\( 75.31^{\circ} \text{C} \)[/tex] and 0.851 atm.

[tex]\[ \text{Cu}_2\text{S} (s) + \text{O}_2 (g) \rightarrow \text{Cu}_2\text{O} (s) + \text{SO}_2 (g) \][/tex]


Sagot :

Sure, let's walk through the solution step by step:

1. Convert the temperature from Celsius to Kelvin:
[tex]\[ T(K) = T(°C) + 273.15 \][/tex]
Given [tex]\( T(°C) = 75.31 \)[/tex]:
[tex]\[ T(K) = 75.31 + 273.15 = 348.46 \, \text{K} \][/tex]

2. Calculate the number of moles of [tex]\( Cu_2S \)[/tex] (Copper(I) sulfide):
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We know the mass of [tex]\( Cu_2S \)[/tex] is 14.29 grams.
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The molar mass of [tex]\( Cu_2S \)[/tex] can be calculated as:
[tex]\[ \text{Molar mass of } Cu_2S = 2 \times 63.55 \, (\text{atomic mass of Cu}) + 32.06 \, (\text{atomic mass of S}) = 159.16 \, \text{g/mol} \][/tex]
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The number of moles is then:
[tex]\[ \text{Moles of } Cu_2S = \frac{\text{mass}}{\text{molar mass}} = \frac{14.29 \, \text{g}}{159.16 \, \text{g/mol}} \approx 0.08978 \, \text{moles} \][/tex]

3. Determine the moles of [tex]\( SO_2 \)[/tex] produced:
From the balanced chemical equation:
[tex]\[ Cu_2S (s) + O_2 (g) \rightarrow Cu_2O (s) + SO_2 (g) \][/tex]
We see that 1 mole of [tex]\( Cu_2S \)[/tex] produces 1 mole of [tex]\( SO_2 \)[/tex]. Therefore:
[tex]\[ \text{Moles of } SO_2 \text{ produced} = \text{Moles of } Cu_2S \approx 0.08978 \, \text{moles} \][/tex]

4. Calculate the volume of [tex]\( SO_2 \)[/tex] gas produced using the Ideal Gas Law:
The Ideal Gas Law is given by:
[tex]\[ PV = nRT \implies V = \frac{nRT}{P} \][/tex]
Where:
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[tex]\( n \)[/tex] = number of moles of [tex]\( SO_2 \)[/tex] = 0.08978 moles,
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[tex]\( R \)[/tex] = Ideal Gas Constant = 0.0821 atm·L/(mol·K),
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[tex]\( T \)[/tex] = temperature in Kelvin = 348.46 K,
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[tex]\( P \)[/tex] = pressure in atm = 0.851 atm.
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Then:
[tex]\[ V = \frac{0.08978 \, \text{moles} \times 0.0821 \, \text{atm·L/(mol·K)} \times 348.46 \, \text{K}}{0.851 \, \text{atm}} \approx 3.018 \, \text{L} \][/tex]

Therefore, the volume of sulfur dioxide ([tex]\( SO_2 \)[/tex]) gas produced is approximately [tex]\( 3.018 \, \text{L} \)[/tex].
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