Question

A gas grill burns Propane (C3H8) in the presence of more than sufficient Oxygen (O2). This reaction produces water vapor and Carbon Dioxide. The temperature and pressure conditions are such that 1 mole of each gas occupies 1 liter of volume.

If 15 liters of Propane are completely consumed, how many grams of Carbon Dioxide are produced?

A. 15 L CO2
B. 45 L CO2
C. 660 g CO2
D. 1980 g CO2
E. None of the Above

Answer 1

Answer on Question #72658, Chemistry / General Chemistry

Question:

A gas grill burns Propane (C₃H₈) in the presence of more than sufficient Oxygen (O₂). This reaction produces water vapor and Carbon Dioxide. The temperature and pressure conditions are such that 1 mole of each gas occupies 1 liter of volume.

If 15 liters of Propane are completely consumed, how many grams of Carbon Dioxide are produced?

A. 15 L CO₂

B. 45 L CO₂

C. 660 g CO₂

D. 1980 g CO₂

E. None of the Above

Solution:

Reaction:

$$\mathrm{C_3H_8} + 5\mathrm{O_2} = 3\mathrm{CO_2} + 4\mathrm{H_2O}$$

Amount of Propane: 15 mol (according to conditions in the task)

Amount of Carbon Dioxide: 15 · 3 = 45 mol

Volume of Carbon Dioxide: 45 L (according to conditions in the task)

Mass of Carbon Dioxide: 45 · 44 = 1980 g

Answer:

D. 1980 g CO₂

Also please note, that answer "B" is correct too. But it shows the volume.

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