Question #215414

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.

Expert's answer

MM(C4H10) = 58.12 g/mol

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

Ideal gas law

"pV=nRT \\\\\n\nV= \\frac{nRT}{p} \\\\\n\np= 121.9 \\;kPa \\\\\n\nT= 26.7 + 273.15 = 299.85 \\;K \\\\\n\nR= 8.314 \\;kPa \\cdot L\/K \\cdot mol \\\\\n\nV = \\frac{0.396 \\times 8.314 \\times 299.85}{121.9}=8.098 \\;L"

Answer: 8.098 L

Learn more about our help with Assignments: Chemistry

## Comments

## Leave a comment