A 10.0 gram sample of a mixture of CH4 and C2H4 reacts with oxygen at 25°C and 1 atm to product carbon dioxide gas and liquid water. If the reaction produces 520 kJ of heat, what is the mass percentage of CH4 in the mixture?
I am all out of tries on the homework, but I would like to know the answer so I can study for my exam!
PLEASE HELP ME, NO ONE ELSE CAN!
I thought I worked this for you a couple of days ago.
I tried to use how you answered for someone else but I got it wrong and used all my tries so I would like the real answer to see where I went wrong!
Thanks for your consideration!
I never worked it through the other day but in setting it up today I think I may have erred in that first post. Give me some time and I'll try to come up with an answer. You can help if you will post what you calculated for dH1, dH2 and the equations.
dH1 i got -890 kJ/mol
dH2 i got -1411 kJ/mol
I came up with 3.11 g CH4 and 6.89 g C2H4.
I checked those answers this way.
The dHrxn for CH4 I found as -890.3 kJ/mol.
and dHrxn for C2H4 as -1410.09 kJ/mol.
Heat produced by 3.11g CH4 is 890.3 kJ/mol x (3.11/16) = 173 kJ.
Heat produced by C2H4 is 1410.09 x (6.89/28) = 347 kJ.
Then heat from CH4 + heat from C2H4 = 173 kJ + 347 kJ = 520 kJ which is what we had in the problem so I feel certain those answers above are correct.
To determine the mass percentage of CH4 in the mixture, we need to set up a balanced chemical equation for the combustion of CH4 and C2H4 to produce CO2 and H2O.
The balanced equation for the combustion of CH4 is as follows:
CH4 + 2O2 → CO2 + 2H2O
The balanced equation for the combustion of C2H4 is as follows:
C2H4 + 3O2 → 2CO2 + 2H2O
Based on the given information, we can use the concept of stoichiometry to solve the problem. We can assume that the reactants have completely reacted, and the given heat of reaction is the heat produced when the given amount of CH4 and C2H4 have burned completely.
First, let's determine the moles of CO2 produced using the given heat release:
The heat produced during combustion is equal to the enthalpy change (ΔH) of the reaction. The standard enthalpy change for the combustion of 1 mole of CH4 is -890.4 kJ.
So, we can use the following equation to calculate the moles of CO2 produced:
520 kJ x (1 mol CO2 / -890.4 kJ) = moles of CO2
Next, let's determine the moles of CH4 and C2H4 used in the reaction:
From the balanced equations, we can see that the ratio of moles of CH4 to moles of CO2 is 1:1, and the ratio of moles of C2H4 to moles of CO2 is 2:2. This means that for every mole of CO2 produced, one mole of CH4 and two moles of C2H4 are consumed.
Since we know the moles of CO2 produced, we can conclude that the same number of moles of CH4 and twice the number of moles of C2H4 were consumed.
Finally, let's calculate the mass of CH4 and C2H4:
To determine the mass percentage of CH4, we need to find the mass of CH4 and the total mass of the mixture. We can calculate these using the molar masses of CH4 and C2H4, which are approximately 16 g/mol and 28 g/mol, respectively.
Mass of CH4 = moles of CH4 x molar mass of CH4
Mass of C2H4 = moles of C2H4 x molar mass of C2H4
Mass percentage of CH4 = (Mass of CH4 / Total mass) x 100
By following these steps and plugging in the values, you should be able to find the mass percentage of CH4 in the mixture.