Chemistry Problem...Thermochemistry?

Question

A 10.0-g sample of a mixture of CH4 and C2H4 reacts with oxygen at 25.0oC and 1.00 atm to produce CO2(g) and H2O(l). If the reaction produces 545 kJ of heat, what is the mass percentage of CH4 in the mixture?

I would appreciate any help, I have tried this so many times...

Neither of these got me the right answer :(

I am STILL not getting the right answer (it's an online homework, so I'll know when I get it). The last person had the right heat of combustion according to my book, though-- -890.23 fr CH4 and -1141.07 for C2H4. I have tried and retried all the methods suggested, and nothing.

Answer 1

I was able to find only combustion heat values at the link specified below so may be you have to substitute other values.
For methane it said the heat is 50.009 MJ/kg, or Q1 = 50.009 kJ/g.
For ethylene the heat is 47.195 Mj/kg, or Q2 = 47.195 kJ/g.
Lets mass of methane is M1 and mass of ethylene is M2, then we have 2 equations:
M1 + M2 = 10
Q1*M1 + Q2*M2 = 545
Solve by substitution: M2 = 10 - M1
Q1*M1 + Q2*(10 - M1) = 545
(Q1 - Q2)*M1 + 10*Q2 = 545
M1 = (545 - 10*Q2)/(Q1 - Q2)
Mass pesentage is
M1/10*100 = 10*M1 = 10*(545 - 10*Q2)/(Q1 - Q2)

Answer 2

First we need to know the heats of reaction for the combustion of both CH4 and C2H4. Since I don't have those numbers, I will call them u and v for now.

Second, we will state that we started off with x grams of CH4, giving us (10-x) grams of C2H4; this means that you had (x/16) mol of CH4 and ((10-x)/28) mol of C2H4.

Finally, based on the heats above, you have (x/16)*u + ((10-x)/28)*v = 545. This is a linear equation, so solve for x.

x/10 = mass percentage of CH4 in the reaction.

Answer 3

CH4 + 2O2 ---> heat of combustion is -891kJ/mol

C2H4 + 3O2 ---> heat of combustion is -1411 kJ/mol

I got these values off the NIST Chemistry WebBook

Convert CH4 value to kJ/g = 55.69
Covert C2H4 value to kJ/g = 50.4

Let x = grams of CH4, therefore grams of C2H4 is 10-x

(55.69) (x) + (50.4) (10-x) = 545

I get x = 7.75 g

Answer 4

First of all, I think you need data for the enthalpy change of combustion for both methane and ethene. I found from the two sources below that ΔHc(methane) = -882.0 kJ/mol and ΔHc(ethene) = -1387.4 kJ/mol. With the information now, we can form two equations to find out the number of moles of methane and ethene used:

Let x and y be the no. of moles of methane and ethene used respectively.

The molar mass of methane is 12 + 4 = 16
The molar mass of ethene is 2(12) + 4 = 28

Since mass = No. of moles x molar mass, the mass of methane used is 16x and mass of ethene used is 28y. So:

16x + 28y = 10 ==> 1

Now, since the ΔHc of both gases are known, we can form an equation relating the 545 kJ of heat they give off.

1 moles of methane burns to give 882.0 kJ of heat.
1 mole of ethene burns to give 1387.4 kJ of heat.

So, x moles of methane gives 882x kJ of heat
So, y moles of ethene gives 1387.4 kJ of heat.

Therefore, 882x + 1387.4y = 545 ==> 2

16x + 28y = 10 ==> 1
882x + 1387.4y = 545 ==> 2

From 1, x = (10-28y) / 16

Substitute this into 2

882[(10-28y) / 16] + 1387.4y = 545
8820 - 24696y + 22198.4y = 8720
-2497.6y = -100
y = 0.04 (approx)

y = 0.04 => x = [10 - 28(0.04)] / 16

x= 0.5549

So, the no. of moles of methane used is 0.5549 moles

Now, we convert his back to mass:

Mass of methane = No. of moles x molar mass
Mass of methane = 0.05549 x 16 = 8.8789 g

Since the total mass of gases is 10.0g, the mass percentage of methane is:

8.8789/10 x 100 = 88.8% (Ans)