How much volume of ethane
would be required to produce 1.12 dm
3 of carbon (IV) oxide on
combustion in sufficient oxygen?
2C2H6(g)+7O2 -> 4CO2(g) + 6H2O�
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To determine the volume of ethane required to produce 1.12 dm^3 of carbon (IV) oxide (CO2) on combustion, we need to take into account the stoichiometry of the balanced equation.
Given the balanced equation:
2C2H6(g) + 7O2 -> 4CO2(g) + 6H2O
We can see that for every 2 moles of ethane (C2H6), we produce 4 moles of carbon (IV) oxide (CO2).
1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, we will need to calculate the number of moles of ethane required and then convert it to the volume in dm^3.
Step 1: Calculate the number of moles of carbon (IV) oxide (CO2)
The volume of CO2 is given as 1.12 dm^3.
We know that 1 mole of any gas occupies 22.4 dm^3 at STP.
So, the number of moles of CO2 produced is 1.12 dm^3 / 22.4 dm^3/mol = 0.05 moles.
Step 2: Calculate the number of moles of ethane required
According to the balanced equation, 2 moles of ethane produce 4 moles of CO2.
So, the number of moles of ethane required is (0.05 moles CO2 * 2 moles C2H6)/4 moles CO2 = 0.025 moles.
Step 3: Convert moles of ethane to volume in dm^3
Since 0.025 moles of ethane produce the desired amount of CO2, we can assume the same proportion for the volume.
1 mole of any gas occupies 22.4 dm^3 at STP.
Hence, 0.025 moles of ethane occupy 0.025 moles * 22.4 dm^3/mol = 0.56 dm^3.
Therefore, approximately 0.56 dm^3 of ethane would be required to produce 1.12 dm^3 of carbon (IV) oxide on combustion in sufficient oxygen.