If you start with 4.00 moles of C3H8 (propane) and 4.00 moles of O2, how many moles of carbon dioxide can be produced?
C₃H₈ + 5 O₂ --> 3 CO₂ + 4 H₂O
4 moles of C₃H₈ would normally produce 12 moles of CO₂ ... but there is insufficient O₂
the O₂ is 1/5 of what is necessary, so the yield will be
... 1/5 * 12 moles
thanks dude
To determine the moles of carbon dioxide (CO2) produced, we need to look at the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the equation, we see that for every 1 mole of propane (C3H8) burned, we produce 3 moles of carbon dioxide (CO2).
Therefore, to find the moles of CO2 produced, we multiply the moles of propane (C3H8) given by the ratio:
4.00 moles C3H8 x (3 moles CO2 / 1 mole C3H8) = 12.00 moles CO2
So, 12.00 moles of carbon dioxide can be produced.
To determine the number of moles of carbon dioxide (CO2) that can be produced when starting with 4.00 moles of C3H8 (propane) and 4.00 moles of O2, we need to balance the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2. Therefore, we need to compare the number of moles of C3H8 and O2 to determine which is the limiting reagent.
To do this, we calculate the stoichiometric ratio of C3H8 to CO2 and O2 to CO2:
- C3H8 to CO2 ratio: 1 C3H8 : 3 CO2
- O2 to CO2 ratio: 5 O2 : 3 CO2
Now, let's calculate the number of moles of CO2 produced from each reactant:
From 4.00 moles of C3H8:
CO2 moles = (4.00 moles C3H8) x (3 moles CO2 / 1 mole C3H8) = 12.00 moles CO2
From 4.00 moles of O2:
CO2 moles = (4.00 moles O2) x (3 moles CO2 / 5 moles O2) = 2.40 moles CO2
We find that the limiting reagent is O2 since it produces fewer moles of CO2 (2.40 mol) compared to the moles produced by C3H8 (12.00 mol).
Therefore, when starting with 4.00 moles of C3H8 and 4.00 moles of O2, the number of moles of carbon dioxide that can be produced is 2.40 moles.