10.0^cm3 of propane was sparked with 60 of oxygen at s.t.p.what is the volume of residu gas after the reaction is cooled
I need the solution to that question
To calculate the volume of the residual gas after the reaction is cooled, we'll need to determine the balanced chemical equation and use the stoichiometry to find the volume.
The balanced chemical equation for the combustion of propane (C3H8) with oxygen (O2) is:
C3H8 + 5O2 -> 3CO2 + 4H2O
According to the equation, 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.
Step 1: Convert the given volume of propane to moles.
The volume of propane given is 10.0 cm3.
Since we are working at standard temperature and pressure (STP), we can use the molar volume of an ideal gas at STP, which is 22.4 L/mol.
Converting cm3 to L, we have 10.0 cm3 = 10.0/1000 L = 0.01 L.
Now, using the molar volume at STP, we can calculate the number of moles of propane:
0.01 L / 22.4 L/mol = 0.0004464 moles of propane.
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of propane (0.0004464 mol) and the moles of oxygen (60 mol).
From the balanced equation, we know that 1 mole of propane reacts with 5 moles of oxygen.
So, the number of moles of oxygen needed to react with 0.0004464 moles of propane is (0.0004464 mol) * (5 mol O2/1 mol C3H8) = 0.002232 mol O2.
Since we have 60 mol of oxygen, which is much greater than the 0.002232 mol needed, the limiting reactant is propane.
Step 3: Calculate the moles of residual gas after the reaction.
From the balanced equation, we can see that the mole ratio between the reactants (propane and oxygen) and the products (carbon dioxide and water) is 1:3.
Since propane is the limiting reactant, it will be fully consumed.
Thus, all the propane will be converted to products, and the moles of residual gas will only include the moles of oxygen that did not react.
The moles of residual gas (oxygen) = (moles of oxygen initially) - (moles of oxygen used in the reaction)
The moles of oxygen used in the reaction = (0.0004464 mol C3H8) * (5 mol O2/1 mol C3H8) = 0.002232 mol O2
The moles of residual gas = 60 mol O2 - 0.002232 mol O2 = 59.997768 mol O2 (rounded to 4 decimal places)
Step 4: Convert the moles of residual gas to volume.
Using the molar volume of an ideal gas at STP (22.4 L/mol), we can calculate the volume of the residual gas:
Volume of residual gas = (moles of residual gas) * (molar volume at STP)
Volume of residual gas = 59.997768 mol O2 * 22.4 L/mol = 1343.99 L (rounded to 2 decimal places)
Therefore, the volume of the residual gas after the reaction is cooled is approximately 1343.99 L.
To find the volume of the residual gas after the reaction is cooled, we first need to determine the balanced chemical equation for the reaction between propane (C₃H₈) and oxygen (O₂).
The balanced chemical equation for the combustion of propane can be written as follows:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
From the balanced equation, we can see that one mole of propane reacts with five moles of oxygen to produce three moles of carbon dioxide (CO₂) and four moles of water (H₂O).
In order to calculate the volume of the residual gas, we need to convert the given volume of propane from cm³ to moles, and then use stoichiometry to determine the corresponding volume of the residual gas.
Step 1: Convert the volume of propane from cm³ to moles
We need to know the molar volume of a gas at STP (Standard Temperature and Pressure). At STP, one mole of any gas occupies a volume of 22.4 L (22,400 cm³). Therefore, to convert cm³ to moles, we divide the given volume (10.0 cm³) by 22.4 cm³/mol.
10.0 cm³ / 22.4 cm³/mol = 0.4464 mol of propane
Step 2: Use stoichiometry to determine the moles of residual gas
From the balanced equation, we can see that one mole of propane produces three moles of carbon dioxide and four moles of water. Therefore, the total moles of products formed (including residual gas) would be 3 + 4 = 7 moles.
Since the oxygen is the limiting reagent (60 moles), all 60 moles of oxygen will be consumed in the reaction, leaving only 7 moles of residual gas.
Step 3: Convert moles of residual gas to volume
To convert the moles of residual gas to volume, we will use the molar volume at STP (22.4 L/mol). Multiplying the number of moles (7 moles) by the molar volume will give us the volume of the residual gas.
Volume of residual gas = 7 moles × 22.4 L/mol = 156.8 L
Therefore, the volume of the residual gas after the reaction is cooled is 156.8 liters.
You have no units for the O2. I assume that is 60 cc.
C3H8 + 5O2 ==> 3CO2 + 4H2O
10 cc C3H8 will need 50 cc O2 to react completely and will produce 30 cc CO2 gas and 40 cc H2O gas (vapor at the combustion temperature). There will be 10 cc O2 remaining un-reacted. After cooling the 40 cc H2O vapor will not be a gas at stp. So the gas volume will be
0 cc C3H8 + 10 cc O2 un-reacted + 30 cc CO2 produced.
I'm not exactly sure what you mean by residual gas. If you mean the volume of the gas at the end of the reaction that will be 10 + 30 = 40 cc. If you mean the volume of the un-reacted O2 gas it will be 10 cc.