Many home barbecues are fueled with propane gas (C3H8). How much carbon dioxide in kilograms is produced upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gal tank)? Assume that the density of the liquid propane in the tank is 0.621g/mL.

I have no idea how to do this and I'm really stressing out on it. I'm thinking about leaving it blank. But I know I have to try.
Help me if you can, thanks!

First, you need to write a balanced equation for the combustion of C3H8. Then, you know how many liters, and you know the density, so you can calculate how many moles of C3H8 you have. Then, you can get the moles of CO2...
I hope this gets you started. Let me know if you're still stuck!!

C3H8 + 5O2 yields 4H2O + 3CO2

The combustion of 18.9 L of propane produces 35.1 kilograms of carbon dioxide.

48 kg

HeLp, A propan cylinder for a grill has an initial mass of 1.297 kg. in order to get the molecular mass of propan you collect 51.50 L of gas by water displacement at 27C. the mass of cylinder after releasing the gas is 1.203kg. if the b pressure is 15.00 psi & the water vapor pressure at 27c is 0.0352atm. I need to use this data to determine molecular mass of the propane and figure out the % error for the mass if the formula is c2h8.

Of course, I'll help you step by step to solve this problem.

Step 1: Write a balanced equation for the combustion of C3H8 (propane):
C3H8 + 5O2 -> 3CO2 + 4H2O

According to the balanced equation, every 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).

Step 2: Calculate the moles of C3H8:
To determine the moles of C3H8, you'll need to convert the volume of propane from liters to moles using the ideal gas law. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Since the problem doesn't provide the temperature or pressure, we can assume standard conditions, which are typically 1 atm pressure and 273.15 K (0 degrees Celsius). The ideal gas constant (R) is 0.0821 L·atm/(mol·K).

First, let's convert the volume of propane from liters to milliliters:
18.9 L x 1000 mL/L = 18,900 mL

Next, we'll convert the volume of propane from milliliters to grams using the given density of 0.621 g/mL:
18,900 mL x 0.621 g/mL = 11,739 g

Now, we'll convert the grams of propane to moles. To do this, we'll use the molar mass of propane, which is the sum of the atomic masses of carbon (C) and hydrogen (H):
3(12.01 g/mol) + 8(1.008 g/mol) = 44.11 g/mol

Moles of C3H8 = 11,739 g / 44.11 g/mol = 265.97 mol

Step 3: Calculate the moles of CO2:
Since the stoichiometry of the balanced equation states that 1 mole of propane reacts to produce 3 moles of CO2, we can use the same conversion factor to determine the moles of CO2.

Moles of CO2 = 265.97 mol x (3 mol CO2 / 1 mol C3H8) = 797.91 mol

Step 4: Convert moles of CO2 to kilograms:
To convert moles of CO2 to kilograms, you'll need to use the molar mass of CO2, which is the sum of the atomic masses of carbon (C) and oxygen (O):
12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol

Mass of CO2 = 797.91 mol x 44.01 g/mol = 35,109.00 g

Finally, convert grams to kilograms:
Mass of CO2 = 35,109.00 g / 1000 = 35.11 kg

Therefore, approximately 35.11 kilograms of carbon dioxide (CO2) are produced upon the complete combustion of 18.9 liters of propane.