Answer to Question #215414 in General Chemistry for Nara

Question

What is the volume of 23.017g of butane gas (C4H10), at 26.7°C and 121.9 kPa? Round your answer to the nearest thousandth.  

Answer 1

MM(C4H10) = 58.12 g/mol

n(C4H10) $= \frac{2}{MM}= \frac{23.017}{58.12}=0.396 \;mol$

Ideal gas law

$pV=nRT \\ V= \frac{nRT}{p} \\ p= 121.9 \;kPa \\ T= 26.7 + 273.15 = 299.85 \;K \\ R= 8.314 \;kPa \cdot L/K \cdot mol \\ V = \frac{0.396 \times 8.314 \times 299.85}{121.9}=8.098 \;L$

Answer: 8.098 L