The balanced equation for the combustion of butane, C4H10, is
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
Calculate the moles of CO2 produced when 2.62 moles of C4H10 are allowed to react with 13.37 moles of O2.
This is another LR problem. Just follow the AlCl3 problem. Post your work if you run into trouble.
From the balanced equation, we can see that the coefficient of C4H10 is 2, which means that for every 2 moles of C4H10, 8 moles of CO2 are produced.
Therefore, we can set up a ratio:
2 moles of C4H10 : 8 moles of CO2
Now, we can use this ratio to calculate the moles of CO2 produced when 2.62 moles of C4H10 react:
(2.62 moles C4H10) x (8 moles CO2 / 2 moles C4H10) = 10.48 moles CO2
So, when 2.62 moles of C4H10 react with 13.37 moles of O2, 10.48 moles of CO2 are produced.
To calculate the moles of CO2 produced, we can use the balanced equation as a conversion factor.
According to the balanced equation, 2 moles of C4H10 react to produce 8 moles of CO2.
So we can set up a ratio:
2 moles C4H10 : 8 moles CO2
To calculate the moles of CO2 produced, we can use the given number of moles of C4H10 and apply the ratio.
Given:
Number of moles of C4H10 = 2.62 moles
Using the ratio, we can calculate the moles of CO2 produced:
2.62 moles C4H10 x (8 moles CO2 / 2 moles C4H10) = 10.48 moles CO2
Therefore, when 2.62 moles of C4H10 react with 13.37 moles of O2, the reaction produces 10.48 moles of CO2.