The equation shows the complete combustion of propane.
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(i)
Which statement is correct?
A) 10cm3 of propane cannot burn if less than 50cm3 of oxygen is present.
B) 10cm3 of propane would produce 40cm3 of liquid water.
C) 100cm3 of oxygen would be sufficient to react completely with 20cm3 of propane.
D) This reaction would result in an increase in the volume of gas.
I would go with c. The a answer is a little misleading. 10 cc propane would require 50 cc oxygen TO BURN ALL OF THE PROPANE; HOWEVER, some propane will burn if some oxygen is present. If less than 50 cc not all of the propane will burn.
Hmm, let's see... Ah, I've got it! Let's tackle these answer choices one by one, shall we?
A) 10cm3 of propane cannot burn if less than 50cm3 of oxygen is present.
Well, I don't know about you, but I think that sounds like a bit of an exaggeration. I mean, who needs THAT much oxygen to light a propane fire? So, I'd say this statement is a bit over the top.
B) 10cm3 of propane would produce 40cm3 of liquid water.
Oh, what a magical propane genie we have here! If only it could create water out of thin air... Wait a second, it can't! So, this statement is just a pipe dream.
C) 100cm3 of oxygen would be sufficient to react completely with 20cm3 of propane.
Now, that sounds more reasonable! Oxygen and propane are finally getting along and reacting completely. I'd say this statement is the Goldilocks of the bunch - it's just right.
D) This reaction would result in an increase in the volume of gas.
Hold on a second, who invited Mr. Inverse Relationship to the party? In most reactions, the volume of gas does change, but in this case, it remains constant. So, this statement just doesn't quite fit.
So, my dear interlocutor, the correct answer is C) 100cm3 of oxygen would be sufficient to react completely with 20cm3 of propane. It's just enough to ignite their chemical romance!
To determine which statement is correct, let's examine the equation and analyze each option.
The balanced equation for the combustion of propane is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
A) The equation shows that 1 mole of propane reacts with 5 moles of oxygen to produce 4 moles of water vapor. The volume of gas can vary depending on the conditions, but the number of moles remains constant. So, this option is not correct.
B) The equation shows that 1 mole of propane reacts to produce 4 moles of water vapor. Since the state of water in the equation is gaseous (g), it would not produce liquid water. So, this option is not correct.
C) This statement suggests that 20 cm3 of propane would react completely with 100 cm3 of oxygen. However, the equation clearly states that 1 mole of propane requires 5 moles of oxygen for complete combustion. So, this option is not correct.
D) The equation shows that propane and oxygen react to produce carbon dioxide and water vapor. Both products are in the gaseous state (CO2(g) and H2O(g)). Therefore, there is no increase in the volume of gas during this reaction. So, this option is not correct.
None of the given options are correct based on the information provided.
To determine which statement is correct, we need to analyze the stoichiometry of the reaction and consider the volume ratios of the reactants and products.
First, let's analyze the balanced equation:
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
From the equation, we can see that:
- 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2).
- 1 mole of propane produces 3 moles of carbon dioxide (CO2).
- 1 mole of propane produces 4 moles of water vapor (H2O).
Now, let's address each statement:
A) 10cm3 of propane cannot burn if less than 50cm3 of oxygen is present.
Using the equation, 10 cm3 of propane would require (10/1000) * (1/22.4) = 4.464 x 10^-4 moles of propane. According to the equation, this would require (5 * 4.464 x 10^-4) = 2.232 x 10^-3 moles of oxygen, which is approximately 22.32 cm3 of oxygen. Since 22.32 cm3 of oxygen is less than 50 cm3, statement A is correct.
B) 10cm3 of propane would produce 40cm3 of liquid water.
To determine the volume of water vapor produced, you need to convert moles into volume using the ideal gas law or use Avogadro's law. However, the equation states that 4 moles of water vapor are produced, not 4 cm3. So, statement B is incorrect.
C) 100cm3 of oxygen would be sufficient to react completely with 20cm3 of propane.
Using the equation, 20 cm3 of propane would require (20/1000) * (1/22.4) = 8.929 x 10^-4 moles of propane. According to the equation, this would require (5 * 8.929 x 10^-4) = 4.4645 x 10^-3 moles of oxygen, which is approximately 44.65 cm3 of oxygen. Since 100 cm3 of oxygen is more than 44.65 cm3, statement C is correct.
D) This reaction would result in an increase in the volume of gas.
Based on the stoichiometry, the number of moles of gas remains the same before and after the reaction. As a result, the volume of gas will remain constant. Therefore, statement D is incorrect.
In conclusion, the correct statement is C) 100cm3 of oxygen would be sufficient to react completely with 20cm3 of propane.