2C2H6+702=4CO2+6H20
how many grams of oxygen are required to completely burn60.0g ethane (C2H6)
show work
Using the above equation how many Lof 02 are required to burn 20.0Lof ethane
thank u soo much this is for a test
See your post later (above). These stoichiometry problems are all the same.
To determine the grams of oxygen required to completely burn 60.0g of ethane (C2H6), we need to use stoichiometry and the balanced equation provided.
1. Start by writing the balanced chemical equation:
2C2H6 + 7O2 → 4CO2 + 6H2O
2. Calculate the molar mass of ethane (C2H6):
(2 x molar mass of carbon) + (6 x molar mass of hydrogen)
= (2 x 12.01 g/mol) + (6 x 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol
3. Use the molar mass of ethane to convert grams to moles:
60.0 g ethane x (1 mol ethane/30.08 g ethane)
= 1.994 mol ethane
4. Use the stoichiometric ratio from the balanced equation to determine the moles of oxygen required:
From the equation, we can see that 2 moles of ethane react with 7 moles of oxygen. Therefore,
1.994 mol ethane x (7 mol oxygen/2 mol ethane)
= 6.978 mol oxygen
5. Finally, convert moles of oxygen to grams:
6.978 mol oxygen x (32.00 g/mol oxygen)
= 223.5 g oxygen
Therefore, 223.5 grams of oxygen are required to completely burn 60.0 grams of ethane.
Now, let's move on to the second question.
To determine the liters of oxygen required to burn 20.0L of ethane, we will follow a similar process:
1. Start with the balanced equation:
2C2H6 + 7O2 → 4CO2 + 6H2O
2. We need to convert 20.0L of ethane to moles. To do this, we need to know the molar volume at a given temperature and pressure. Let's assume standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere. At STP, 1 mol of any ideal gas occupies 22.4 Liters.
So, to calculate moles of ethane:
20.0L ethane x (1 mol/22.4 L)
= 0.893 mol ethane
3. Now, using the stoichiometric ratio from the balanced equation, determine the moles of oxygen:
From the equation, we can see that 2 moles of ethane react with 7 moles of oxygen. Therefore,
0.893 mol ethane x (7 mol oxygen/2 mol ethane)
= 3.119 mol oxygen
4. But we need to convert moles of oxygen to liters at STP:
Using the same molar volume at STP (22.4L/mol), we can multiply the moles of oxygen by the molar volume:
3.119 mol oxygen x (22.4 L/mol)
= 69.87L oxygen
Therefore, approximately 69.87 liters of oxygen are required to burn 20.0 liters of ethane.