Question

Balance the equation. Then, determine the corresponding amount in moles of each of the other reactants and products

NH3

4 mol N2

4.5 mol O2

Answer 1

shame on me for not knowing...some hints on what are the products of the reaction?

Answer 2

NH3 + O2 > N2 + H2O

Answer 3

do you know how to do atom inventory?

Answer 4

no :(

Answer 5

very briefly: we count the number of atoms for each element for the left hand side. then we do the same for the right hand side. and we compare the numbers.

for example: NH3 + O2 > N2 + H2O

on the left: N: 1, H: 3, O: 2
on the right: N: 2, H: 2, O: 1

the numbers don't match so it's not balanced.

Answer 6

Ah, then what

Answer 7

We have to balance the chemical equation
\[\text{NH}_3 + \text O_2 \to \text N_2 + \text H_2\text O\]
Lets look at the Nitrogen first;
We saw that according to gsoda's inventory, there are twice as many Nitrogen atoms in the products than in the reactants, so twice the number of ammonia is required.
\[2\text{NH}_3 + \text O_2 \to \text N_2 + \text H_2\text O\]
the Nitrogen in now balanced,
but we have more Hydrogen on the left than we do on the right so ...
\[2\text{NH}_3 + \text O_2 \to \text N_2 + 3\text H_2\text O\]
now the Nitrogen and the Hydrogen are balanced,
we just have to balance the Oxygen.
\[2\text{NH}_3 + \tfrac32\text O_2 \to \text N_2 + 3\text H_2\text O\]
however the stoichiometric ratios must be natural numbers ,
so we have to multiply every term by 2.
\[\boxed{4\text{NH}_3 + 3\text O_2 \to 2\text N_2 + 6\text H_2\text O}\]____

If you make an atom inventory now, you will see the numbers of each atom in the reaction is the same on both side. The equation is now balanced. The coefficients of the chemical in the reaction are in the stoichiometric ratios of the reaction.

This means;
that exactly four atoms of ammonia react with three atoms of Oxygen to form two Nitrogen molecules and six water molecules.

or equivalently

Four moles of ammonia react with three moles of Oxygen to form two moles of Nitrogen and six moles of water .