Question: The psychoactive drug methamphetamine which is sold as the prescription medication Desoxyn C10H15N, undergoes a series of reactions in the body; the net result of these reactions is the oxidation of solid methamphetamine by oxygen gas to produce carbon dioxide gas, liquid water, and an aqueous solution of urea, CH4N2O. Write the balanced equation for this net reaction.
I can't seem to balance this equation. If someone could clarify the steps needed to solve this problem I'd appreciate it!
Question H21 (6th Edition)
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 88
- Joined: Fri Sep 28, 2018 12:25 am
Re: Question H21 (6th Edition)
For the unbalanced equation we have:
C10H15N + O2 --> CO2 + H2O + CH4N20
I started with balancing N first:
2C10H15N + O2 --> CO2 + H2O + CH4N20
Then H:
2C10H15N + O2 --> CO2 + 13H2O + CH4N20
Then C:
2C10H15N + O2 --> 19CO2 + 13H2O + CH4N20
Finally O:
2C10H15N + 26O2 --> 19CO2 + 13H2O + CH4N20
C10H15N + O2 --> CO2 + H2O + CH4N20
I started with balancing N first:
2C10H15N + O2 --> CO2 + H2O + CH4N20
Then H:
2C10H15N + O2 --> CO2 + 13H2O + CH4N20
Then C:
2C10H15N + O2 --> 19CO2 + 13H2O + CH4N20
Finally O:
2C10H15N + 26O2 --> 19CO2 + 13H2O + CH4N20
-
- Posts: 35
- Joined: Fri Sep 28, 2018 12:24 am
Re: Question H21 (6th Edition)
(1) : Balance N first because it appears the less within the equation.
2C10H15N + O2 --> CO2 + H2O + CH4N2O
(2): Balance H, there are 30 H on the left side and a total of 6 H on the right, leaving the (4 H) alone, that would be 30-4= 26/2= 13 as your coefficient for H2O
2C10H15N + O2 --> CO2 + 13H2O + CH4N2O
(3): Balance C, there are 20 C on the left side and on the right there is a total of 2 C, if we leave CH4N2O alone then 20-1=19 (1 coming from the one carbon)-- as your coefficient for CO2
2C10H15N + O2 --> 19CO2 + 13H2O + CH4N2O
(4): Balance O which appears the most, there are only 2 on the left side and 38+13+1= 52 O on the right side 52/2=26 as your coefficient for O2
2C10H15N + 26O2 --> 19CO2 + 13H2O + CH4N2O
2C10H15N + O2 --> CO2 + H2O + CH4N2O
(2): Balance H, there are 30 H on the left side and a total of 6 H on the right, leaving the (4 H) alone, that would be 30-4= 26/2= 13 as your coefficient for H2O
2C10H15N + O2 --> CO2 + 13H2O + CH4N2O
(3): Balance C, there are 20 C on the left side and on the right there is a total of 2 C, if we leave CH4N2O alone then 20-1=19 (1 coming from the one carbon)-- as your coefficient for CO2
2C10H15N + O2 --> 19CO2 + 13H2O + CH4N2O
(4): Balance O which appears the most, there are only 2 on the left side and 38+13+1= 52 O on the right side 52/2=26 as your coefficient for O2
2C10H15N + 26O2 --> 19CO2 + 13H2O + CH4N2O
Return to “Balancing Chemical Reactions”
Who is online
Users browsing this forum: No registered users and 1 guest