Sure, let's break it down step by step.
### Part (a): Calculate the volume of oxygen needed to burn 50 cm³ of butane at STP.
First, let's understand the given balanced chemical equation:
[tex]\[2 \text{C}_4\text{H}_{10} (g) + 13 \text{O}_2 (g) \rightarrow 8 \text{CO}_2 (g) + 10 \text{H}_2\text{O} (g)\][/tex]
From this equation, we can see that for every 2 volumes of butane ([tex]\(\text{C}_4\text{H}_{10}\)[/tex]), 13 volumes of oxygen ([tex]\(\text{O}_2\)[/tex]) are needed.
Now, you are given [tex]\(50 \, \text{cm}^3\)[/tex] of butane ([tex]\(\text{C}_4\text{H}_{10}\)[/tex]).
1. Determine the volume ratios from the balanced equation:
[tex]\[2 \, \text{C}_4\text{H}_{10} : 13 \, \text{O}_2\][/tex]
2. Calculate the volume of oxygen needed using the ratio:
[tex]\[
\text{Volume of } \text{O}_2 = 50 \, \text{cm}^3 \times \left(\frac{13}{2}\right) = 325 \, \text{cm}^3
\][/tex]
So, the volume of oxygen needed to burn [tex]\(50 \, \text{cm}^3\)[/tex] of butane at STP is [tex]\(325 \, \text{cm}^3\)[/tex].
### Part (b): Calculate the volume of oxygen at 15°C and 289 mmHg pressure.
Next, we need to adjust the volume of oxygen from STP conditions to the given conditions of 15°C and 289 mmHg. We'll use the Ideal Gas Law for this:
[tex]\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\][/tex]
Where:
- [tex]\(P_1\)[/tex] is the initial pressure at STP, which is [tex]\(760 \, \text{mmHg}\)[/tex].
- [tex]\(V_1\)[/tex] is the initial volume of oxygen at STP, which is [tex]\(325 \, \text{cm}^3\)[/tex].
- [tex]\(T_1\)[/tex] is the initial temperature at STP, which is [tex]\(273.15 \, \text{K}\)[/tex].
- [tex]\(P_2\)[/tex] is the final pressure, which is [tex]\(289 \, \text{mmHg}\)[/tex].
- [tex]\(V_2\)[/tex] is the final volume of oxygen at the given conditions.
- [tex]\(T_2\)[/tex] is the final temperature, which is [tex]\(15^{\circ} \text{C} + 273.15 = 288.15 \, \text{K}\)[/tex].
Now, using the proportionality from the Ideal Gas Law, solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}
\][/tex]
[tex]\[
V_2 = \frac{760 \times 325 \times 288.15}{289 \times 273.15} \approx 901.6 \, \text{cm}^3
\][/tex]
Therefore, the volume of oxygen at [tex]\(15^{\circ} \text{C}\)[/tex] and [tex]\(289 \, \text{mmHg}\)[/tex] is approximately [tex]\(901.6 \, \text{cm}^3\)[/tex].
