C3H8 + 5O2 ==> 3CO2 + 4H2O
When using gases at the same T and P, one can take a shortcut and use liters as if they were mols. For the problem we simply convert L of one thing into L of the other.
0.7L C3H8 x (5 mols O2/1 mol C3H8) = ? L O2 required.
The other parts are done the same way.
Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).
Write a balanced equation for this reaction.
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of oxygen will be required to completely burn 0.700 L of propane gas?
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of carbon dioxide gas will be produced in the reaction?
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of water vapor will be produced in the reaction
2 answers
1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
1 answer