Question

balance this equation fast

Answer 1

Hope you know all the formula:
Copper=Cu
Nitric acid= HNO₃
Copper nitrate=Cu(NO₃)₂
Nitric oxide= NO
Water= H₂O

according to the given reaction:

Cu + HNO₃ ⇒ Cu(NO₃)₂ + NO +  H₂O

Now to balance it:
#Cu in LHS and RHS same ,so no problem.

# N in LHS=1
  N in RHS =3
Multiply the N containing compound in LHS by 3

We get:
Cu + 3 HNO₃ ⇒ Cu(NO₃)₂ + NO +  H₂O

O in LHS=9 (odd)
O in RHS =8 , again problem :(
so change 3HNO₃ to 4HNO₃
now O in LHS= 12, to balance it with RHS , multiply H₂O by 4

we get:
Cu + 4 HNO₃ ⇒ Cu(NO₃)₂ + NO +  4 H₂O

H is not balanced 
H in LHS = 4
H in RHS =8

change 4 HNO₃ to 8HNO₃ to balance it with RHS we get,

Cu + 8 HNO₃ ⇒ Cu(NO₃)₂ + NO +  4 H₂O

hmm now whats left ?

N in LHS= 8
N in RHS=3 (odd)
To amke N balanced in LHS and RHS ,multiply Cu(NO₃)₂ by 3 and NO by 2

we get,
 Cu + 8 HNO₃ = 3 Cu(NO₃)₂ + 2 NO + 4 H₂O
Now, only Cu is LEFT to balance in LHS, multiply Cu in LHS by 3. we get,

3Cu + 8 HNO₃ ⇒ 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

Its finally done.

Lets check.
# Cu
LHS=3             RHS=3

#H
LHS=8            RHS=8

#N
LHS=8            RHS=8

#O
LHS=24          RHS=24