Answer:
See below
Explanation:
2C2H6 + 7O2 = 4CO2 + 6H2O
2C2H2 + 5O2 = 4CO2 + 2H2O
A lot of this is trial and error. I start from left to right. For the first reaction, you had
C2H6 + O2 = CO2 + H2O
You have 2 carbons on the left so you can try adding a 2 in front of the carbons on the right and keep going from there. Now you have
C2H6 + O2 = 2CO2 + H2O
You now have 2 C on each side. C2 is the same as 2C.
The next step is to look at the hydrogens on the left. You have 6 so you need 6 on the right. You can add a 3 in front of the H on the right. The H already has a subscript of a 2 so having the 3 in front will now make it 6H. You now have this
C2H6 + O2 = 2CO2 + 3H2O
So your C are equal (2 on each side) and your H is equal (6 on each side). All that is left is the O. So far you have 2 on the left but you have 7 on the right (4 from the 2CO2 and 3 more from 3H2O). Since you have an even number on one side and an odd number on the other, there is no way to make this work. So it's time to start over.
My next step was to put a 2 in front of C2H6 and keep going as we did above. Start with this
2C2H6 + O2 = CO2 + H2O
You have 4 C on the left. So you need to add a 4 in front of CO2 to get 4 C on the right. Now you have this
2C2H6 + O2 = 4CO2 + H2O
Now on the left you have 12 H (2 in front times the 6 subscript on the H). You need to have 12 on the right. We can get that by putting a 6 in front of the H2O ( again the 6 in front times the 2 subscript of the H is 12 total). You have this now
2C2H6 + O2 = 4CO2 + 6H2O
You now have 4 C on both sides, 12 H on both sides. All that is left now is the O.
On the right side, you have a total of 14 ( 8 from the 4CO2 and 6 from 6H2O). So you'd need to add 7 in front of the O2 to make it 14.
You're done. 4 C on both sides, 12 H on both sides and 14 O on both sides.
The second equation I just did the same process.
