always start with the most difficult formula
C12H22O11 + O2>>CO2 + H20
ok, start then with the 12c
C12H22O11 + O2>>12CO2
Next, H 22
C12H22O11 + O2>>12CO2 + 11H2O
now you have to go back and forth on the O. Since the products are balanced, add the O in th products. 12*2+11=35
In the sucrose formula, you have 11, so that means you need 24O
C12H22O11 + 12O2>>12CO2 + 11H2O
done.
next, start with the CH3OH, the C
CH3OH +O2>>CO2
balanced, then the H
CH3OH + O2>>CO2+ 2H2O
Now, the O THe right side is OK on C and H. Lets start there. On the right side 4, so we need 4 on the left.
CH3OH + 3/2 O2>>CO2 + 2H2O
and it is balanced. Well, some folks in the US get all unbalanced about fraction coefficents (coefficents represent mole ratios, so it is ok. But some folks think like John Dalton did, that the coefficents represent individual molecules, and fractions are not nice). So to placate those individuals, if we get rid of the 3/2..multipy all by 2
2CH3OH + 3O2>>2CO2+4H2O
Balance the following equations.
3. C12H22O11 + O2 = CO2 + H2O
Answer: C12H22O11 + 12 O2 = 12 CO2 + 11 H2O
8. CH3OH + O2 = CO2 + H2O
Answer: 2 CH3OH + 3 O2 = 2 CO2 + 4 H2O
I balanced all the others just fine. But I seem to be having trouble with these two. Can someone show me the steps to get to the answer?
Thank you very much.
1 answer
1 answer