30.0 g Ethane burns in excess oxygen
1) Write the balanced Combustion reaction
2) Determine the number of moles of oxygen required for complete combustion
3) What mass of carbon dioxide is produced?
2C2H6 + 7O2 ==> 4CO2 + 6H2O
b. moles C2H6 = g/molar mass = 30.0/30 = 1
1 mol C2H6 requires 7 moles O2 (look at the coefficients in the balanced equation).
c. 1 mol C2H6 will produce 2 mols CO2. Again, look at the equation. 1 mol CO2 x (4 mols CO2/2 mols C2H6) = 2 mols CO2
Then, g CO2 = mols CO2 x molar rmass CO2
Hm
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Wait how
1) To write the balanced combustion reaction, we need to know the molecular formula for ethane (C2H6) and the products of combustion (carbon dioxide and water):
C2H6 + O2 -> CO2 + H2O
2) The balanced equation tells us that the ratio of ethane to oxygen is 1:3. This means that for every 1 mole of ethane that burns, we need 3 moles of oxygen. Therefore, to determine the number of moles of oxygen required for complete combustion, we multiply the moles of ethane by the ratio:
1 mole of C2H6 * (3 moles of O2 / 1 mole of C2H6) = 3 moles of O2
So, 3 moles of oxygen are required for complete combustion.
3) From the balanced equation, we can see that the ratio of carbon dioxide to ethane is 1:2. This means that for every 1 mole of ethane burned, 2 moles of carbon dioxide are produced. To determine the mass of carbon dioxide produced, we first need to calculate the number of moles of ethane:
Given mass of ethane = 30.0 g
Molar mass of ethane (C2H6) = 30.0 g/mol
Number of moles of ethane = (Given mass of ethane / Molar mass of ethane)
= (30.0 g / 30.0 g/mol)
= 1 mole
Since 1 mole of ethane produces 2 moles of carbon dioxide, the mass of carbon dioxide produced can be calculated using the molar mass of carbon dioxide (CO2):
Molar mass of CO2 = 44.0 g/mol
Mass of carbon dioxide produced = (Number of moles of carbon dioxide * Molar mass of CO2)
= (2 moles * 44.0 g/mol)
= 88.0 g
Therefore, 88.0 g of carbon dioxide is produced.