Question

butanol thermodynamics
volume of co₂ produced
25.0 g c₄h₉oh burns in excess oxygen at
225 °c and 745 mm hg, forming 1.35 moles
co₂.

c₄h₉oh(l) + 6o₂(g) → 4co₂(g) + 5h₂o(g)

use the following equations and constants
to help solve this problem.

pv = nrt r = 62.4 \frac{l·mm hg}{mol·k}

what volume of co₂ forms during the
combustion of the butanol under these
conditions?
0.0393 l
22.4 l
56.3 l
25.4 l

Explanation:

Step1: Identify known values

We know the number of moles of \( CO_2 \) (\( n = 1.35 \) mol), pressure (\( P = 745 \) mm Hg), temperature (\( T = 225 + 273.15 = 498.15 \) K), and the gas constant (\( R = 62.4 \frac{L\cdot mm\ Hg}{mol\cdot K} \)). We use the ideal gas law \( PV = nRT \) and solve for \( V \).

Step2: Rearrange the ideal gas law for volume

From \( PV = nRT \), we can rearrange to solve for \( V \):
\( V=\frac{nRT}{P} \)

Step3: Substitute the known values into the formula

Substitute \( n = 1.35 \) mol, \( R = 62.4 \frac{L\cdot mm\ Hg}{mol\cdot K} \), \( T = 498.15 \) K, and \( P = 745 \) mm Hg into the formula:
\( V=\frac{1.35\ mol\times62.4\frac{L\cdot mm\ Hg}{mol\cdot K}\times498.15\ K}{745\ mm\ Hg} \)

Step4: Calculate the numerator first

First, calculate the numerator:
\( 1.35\times62.4\times498.15 \)
\( 1.35\times62.4 = 84.24 \)
\( 84.24\times498.15 \approx 84.24\times498 \approx 41951.52 \)

Step5: Divide by the pressure

Now divide by the pressure (\( 745 \) mm Hg):
\( V=\frac{41951.52}{745} \approx 56.3 \) L

Answer:

56.3 L