Question

Niacin (`C_6NH_5O_2`) reacts with `O_2` to produce `CO_2`, `H_2O`, and `NO_2`. How would you write and balance the equation for this reaction?

Answer 1

`4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2`

Explanation:

After adjusting the no of atoms of 6C,1N and 5 H in RHS with LHS ,we get total `(14+5/2)` O at RHS.
To adjust the no of O atom in
LHS we can write `(6+5/4)`before `O_2` in LHS and the balanced equation becomes.

`C_6NH_5O_2+(6+5/4)O_2->6CO_2+5/2H_2O+NO_2`

Finally multiplying by 4 on both sides to remove fraction,we get the balanced equation as follows.

`4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2`