Propane gas 1C3H82 is often used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0 L and a pressure of 4.50 atm at 25.0 °C? The equation for the combustion of propane is:
C3H81g2 + 5 O21g2¡3 CO21g2 + 4 H2O1l2
(LO 10.4, 10.5)
(a) 61.8 L (b) 186 L
(c) 20.6 L (d) 2.21 * 103 L

Question

Propane gas 1C3H82 is often used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0 L and a pressure of 4.50 atm at 25.0 °C? The equation for the combustion of propane is: 
C3H81g2 + 5 O21g2¡3 CO21g2 + 4 H2O1l2
(LO 10.4, 10.5)
(a) 61.8 L (b) 186 L
(c) 20.6 L (d) 2.21 * 103 L

Accepted Answer

Hello everyone today, we are being given the following question. Butane gas is used as fuel for portable gas stoves, calculate the volume in liters of C. 02 gas or carbon dioxide gas formed at standard temperature pressure when butane is combusted in a 10 liter reaction vessel with a pressure of 3.15 atmospheres at 25 degrees Celsius. The balanced reaction equation is as follows. So the first thing, the very first thing that we want to do is calculate our moles of beauty to figure out how much we're working with. And so before we do that, we have to recall our gas law. So recall our ideal gas long, which states that the pressure times the volume equals the number of moles times a gas costs in our times our temperature T. And so we want to rearrange this to solve for moles or in and so we get that N is equal to our pressure times volume over r gas constant, R times r temperature T. And so before we plug in our values, we have to make sure that we have all of our variables that we need. So for our pressure It was that was 3.15 atmospheres, Our Volume 10 L. Our temperature Is going to be 25°C. However, we must convert this to Kelvin and to do that, we simply add to 73.15 to get to 0.15 kelvin R. Gas constant are simply going to be 0.08, 206 leaders times atmospheres over moles, times kelvin. So now that we have the variables that we need. We can plug them into this equation here. So far pressure we set was 3.15 atmospheres. Our vol 10 l r gas constant is that point 08206 Leaders Times atmospheres over moles times kelvin Times are temperature just to 98.15 Kelvin. Once our units cancel out we are left with in or the number of moles being 1.2875 moles of our butane. We're not done yet. We must then calculate our moles of carbon dioxide. And so what that's going to look like is that what that is going to look like? The moles for butane that we can that we sold for our 1.28 75 malls of beauty. And so to get from most of Butane, two moles of carbon dioxide. We have to use our multiple ratio and we use our chemical reaction for that. And so we're actually gonna use the coefficients. And so we can say that for every eight moles of carbon dioxide Given by the coefficient of eight in front of the carbon dioxide we have two moles given by the two coefficient in front of our beauty, we have two moles of butane On most of butane canceled out. And then we are left with our 5. moles of carbon dioxide. Next we're going to recall what it means to be at standard temperature and pressure. And this just states that we have Our temperature being 273.15 Kelvin and our pressure being at one atmospheres. And so now that we have all of our variables, we can now solve for our volume in leaders. So we have our ideal gas law that we determined in step two and when we rearrange that to solve the volume, we get that volume is equal to our number of moles that we have, which will be solved for for CO2 was 5.14996 Moles of CO two. That's going to be multiplied by our gas constant R which is point oh 8 to 06 Leaders times atmospheres over moles, times kelvin. And that's gonna be multiplied by our temperature at standard temperature and pressure, which is to 73.15. That's all going to be divided by our one atmospheres. And so essentially we're going to get a volume of L. And this actually corresponds to answer choice C. And with that we have answered the question overall, I hope that this helped and until next time

Verified Solution

Key Concepts

Stoichiometry of Combustion Reactions

Ideal Gas Law

Conditions of STP