Question

Consider the above unbalanced equation. What volume of CO2 is produced at 270. mm Hg and 38.5°C when 0.820 g of C4H8 reacts with excess O2? Use molar masses with at least as many significant figures.

Answer 1

Answer:

The volume of CO2 is 4.20 L

Explanation:

Step 1: Data given

Pressure = 270 mm Hg =  260 /760 = 0.355263 atm

Temperature : 38.5 °C = 311.65 K

Mass of C4H8 = 0.820 grams

Step 2: The balanced equation

C4H8 + 6O2 → 4CO2 + 4H2O

Step 3: Calculate moles C4H8

Moles C4H8 = mass C4H8 / molar mass C4H8

Moles C4H8 = 0.820 grams / 56.11 g/mol

Moles C4H8 = 0.0146 moles

Step 4: Caalculate moles CO2

For 1 mol C4H8 we need 6 moles O2 to produce 4 moles CO2 and 4 moles H2O

For 0.0146 moles we'll have 4*0.0146 = 0.0584 moles CO2

Step 6: Calculate Volume CO2

p*V = n*R*T

V = (n*R*T) /p

⇒ with V = the volume of CO2 = TO BE DETERMINED

⇒ with n = the moles of CO2 = 0.0584 moles

⇒ with R = the gas constant = 0.08206 L*atm / mol*K

⇒ with T = The temperature = 311.65 K

⇒ with p = the pressure = 0.355263 atm

V = (0.0584 * 0.08206 * 311.65) / 0.355263

V = 4.20 L

The volume of CO2 is 4.20 L