To determine the volume of oxygen gas required to react with 13.5 grams of copper (I) sulfide [tex]\(Cu_2S\)[/tex] at a temperature of [tex]\(30.7^{\circ} \text{C}\)[/tex] and pressure of [tex]\(4.6 \text{ atm}\)[/tex], we follow these steps:
1. Convert the temperature to Kelvin:
[tex]\[
T(K) = T(°C) + 273.15 = 30.7 + 273.15 = 303.85 \text{ K}
\][/tex]
2. Calculate the moles of [tex]\(Cu_2S\)[/tex]:
The molar mass of [tex]\(Cu_2S\)[/tex] is 159.16 g/mol.
[tex]\[
\text{Moles of } Cu_2S = \frac{\text{mass}}{\text{molar mass}} = \frac{13.5 \text{ g}}{159.16 \text{ g/mol}} \approx 0.08482 \text{ mol}
\][/tex]
3. Determine the stoichiometry factor:
From the balanced chemical equation:
[tex]\[
Cu_2S(s) + O_2(g) \rightarrow Cu_2O(s) + SO_2(g)
\][/tex]
It is clear that 1 mole of [tex]\(Cu_2S\)[/tex] reacts with 1 mole of [tex]\(O_2\)[/tex]. Thus, the stoichiometry factor is 1.
4. Calculate the moles of [tex]\(O_2\)[/tex] needed:
[tex]\[
\text{Moles of } O_2 \text{ needed} = \text{moles of } Cu_2S \times \text{stoichiometry factor} = 0.08482 \text{ mol} \times 1 = 0.08482 \text{ mol}
\][/tex]
5. Use the ideal gas law to calculate the volume of [tex]\(O_2\)[/tex]:
The ideal gas law is given by [tex]\(PV = nRT\)[/tex]. Solving for the volume [tex]\(V\)[/tex]:
[tex]\[
V = \frac{nRT}{P}
\][/tex]
Where:
- [tex]\(n\)[/tex] is the moles of [tex]\(O_2\)[/tex]: 0.08482 mol
- [tex]\(R\)[/tex] is the ideal gas constant: 0.0821 L·atm/(mol·K)
- [tex]\(T\)[/tex] is the temperature in Kelvin: 303.85 K
- [tex]\(P\)[/tex] is the pressure in atm: 4.6 atm
Plugging in the values:
[tex]\[
V = \frac{0.08482 \text{ mol} \times 0.0821 \text{ L·atm/(mol·K)} \times 303.85 \text{ K}}{4.6 \text{ atm}}
\][/tex]
Simplifying the expression:
[tex]\[
V \approx 0.45999 \text{ L}
\][/tex]
Hence, the volume of oxygen gas required is approximately [tex]\(0.460 \text{ L}\)[/tex] (rounded to three significant figures).
