When 2.75 dm3 of O2 react with an excess of glucose (C6H12O2), according to the reaction below, what volume of carbon dioxide will be produced?
6O2(g) + C6H12O6(s) ---> 6H2O(g) + 6CO2(g)
6O2(g) + C6H12O6(s) ---> 6H2O(g) + 6CO2(g)
When dealing with gases one may take a shortcut and use volume as mols; therefore, 2.75 cm^3 x (6 mols CO2/6 mols O2) = 2.75 cm^3 CO2
To find the volume of carbon dioxide produced, we need to use stoichiometry. The balanced equation tells us that 6 moles of oxygen react with 1 mole of glucose to produce 6 moles of carbon dioxide.
Given: 2.75 dm3 of oxygen
First, we need to convert the volume of oxygen from dm3 to moles, using the ideal gas law:
n = V / Vm
Where:
n = number of moles
V = volume in dm3
Vm = molar volume (which is 22.4 dm3/mol at standard temperature and pressure)
n = 2.75 dm3 / 22.4 dm3/mol
n ≈ 0.123 mol
Now that we have the number of moles of oxygen, we can determine the number of moles of carbon dioxide produced using the stoichiometric ratio from the balanced equation:
6 moles of oxygen react with 6 moles of carbon dioxide
Therefore, 0.123 mol of oxygen will produce 0.123 mol of carbon dioxide.
Finally, we can convert the number of moles of carbon dioxide to volume using the ideal gas law:
V = n * Vm
V = 0.123 mol * 22.4 dm3/mol
V ≈ 2.75 dm3
So, when 2.75 dm3 of oxygen reacts with an excess of glucose, approximately 2.75 dm3 of carbon dioxide will be produced.
To find the volume of carbon dioxide (CO2) produced when 2.75 dm3 of oxygen (O2) react with an excess of glucose (C6H12O6), we can use the balanced equation that relates the molar ratio between O2 and CO2.
The balanced equation is:
6O2(g) + C6H12O6(s) -> 6H2O(g) + 6CO2(g)
From the equation, we can see that the molar ratio between O2 and CO2 is 6:6, meaning for every 6 moles of O2, we will have 6 moles of CO2 produced.
To calculate the volume of CO2 produced, we need to determine the number of moles of O2 first. We can use the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we are given the volume of O2 (2.75 dm3) and assuming the reaction takes place at standard temperature and pressure (STP), which is 0°C (273K) and 1 atmosphere (atm), we can use the inverse of the ideal gas equation to find the number of moles of O2:
n(O2) = PV / RT = (1 atm) * (2.75 dm3) / (0.0821 L*atm/mol*K) * (273 K)
Note: We convert dm3 to liters (L) by multiplying by 1 (since 1 dm3 = 1 L).
Once we know the number of moles of O2, we can use the molar ratio from the balanced equation to determine the number of moles of CO2 produced. Since the molar ratio is 6:6 (O2:CO2), the number of moles of CO2 will be the same as the number of moles of O2.
Finally, to convert the number of moles of CO2 to volume, we can use the inverse of the ideal gas equation again:
V(CO2) = n(CO2) * RT / P = n(O2) * RT / P
Now we can substitute the values we calculated to find the volume of CO2 produced.
It's important to note that this is a theoretical calculation assuming ideal conditions, and in practice, other factors may affect the actual volume of CO2 produced.