How do you balance #NH_3(l) + O_2(g) - NO(g) + H_2O(l)#?

Question

How do you balance #NH_3(l) + O_2(g) -> NO(g) + H_2O(l)#?

Caleb Adams · Accepted Answer

#"4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(l)#

The imbalanced equation given is

#"NH"_3(l) + "O"_2(g) -> "NO"(g) + "H"_2"O"(l)# As a thumb rule #"H" and "O"# are left for the last. As we see that other than these two atoms there is only one atom, #"N"#. That atom is already balanced.

We start with #"H" and "O"#. We observe that out of these two, hydrogen has more number of atoms in the equation. So we take it first.

Let the balanced equation be: (notice that number of N atoms have been kept balanced) #x"NH"_3(l) + y"O"_2(g) -> x"NO"(g) + b"H"_2"O"(l)#

To balance hydrogen atoms we multiply ammonia molecule with 2 (set #x=2#) on reactants side and water molecule by 3 (set #b=3#) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set #x=2#. The equation looks like

#"2NH"_3(l) + y"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)#

For the remaining unbalanced element: Total number of oxygen atoms on products side becomes #2+3=5#. Which gives us number of oxygen molecules as #y=5/2# on the reactants side.

#"2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l)# We know that #y# needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

#2xx("2NH"_3(l) + 5/2"O"_2(g) -> 2"NO"(g) + 3"H"_2"O"(l))# #"4NH"_3(l) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(l)#

Olivia Anton · Answer

undefined 4 NH₃(l) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l)

How do you balance #NH_3(l) + O_2(g) -&gt; NO(g) + H_2O(l)#?

How do you balance #NH_3(l) + O_2(g) -> NO(g) + H_2O(l)#?