# Balancing Chemical Equations

Balancing Chemical Reactions - Real Chemistry

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Let us use a double displacement reaction of Lead (II) Nitrate and Potassium Chromate to produce Lead (II) Chromate and Potassium Nitrate to practice balancing an equation.

We begin with the base equation provided in the question.

$P b {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + {K}_{2} C r {O}_{4 \left(a q\right)} \rightarrow P b C r {O}_{4 \left(s\right)} + K N {O}_{3 \left(a q\right)}$

Looking at the the atom inventory

Reactants
$P b = 1$
$N {O}_{3} = 2$
$K = 2$
$C r {O}_{4} = 1$

Products
$P b = 1$
$N {O}_{3} = 1$
$K = 1$
$C r {O}_{4} = 1$

We can see that the $K$ and $N {O}_{3}$ are imbalanced.

If we add a coefficient of 2 in front of the $K N {O}_{3}$ this will balance the equation.

$P b {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + {K}_{2} C r {O}_{4 \left(a q\right)} \rightarrow P b C r {O}_{4 \left(s\right)} + 2 K N {O}_{3 \left(a q\right)}$

Note that I leave the polyatomic ions$N {O}_{3}$ and $C r {O}_{4}$ together when they appear on both sides of the equation seeing them as one unit not separate elements.

I would watch the following videos to understand the process of balancing equations more in depth.

F2f

• Conservation of matter is the law. You can also call it the conservation of mass.

When we balance an equation, we determine the ratio of reactants to products which allows for the total number of atoms of reactants to match the number of atoms of the products. Since the type of atoms does not change (nuclear processes are a different story) and the number of atoms stays that same, the total mass that goes into the chemical change will match the mass that comes out after the change.

Here is an example:

2${H}_{2}$ + ${O}_{2}$ -> 2${H}_{2} O$

OR

H-H + H-H + O=O -> H-O-H + H-O-H

There are 4H atoms before and after the reaction (each with a mass of 1 amu)
There are 2O atoms before and after the reaction (each with a mass of 16 amu)

The total mass before the reaction is 4x1 + 2x16 = 36amu
The total mass after the reaction is 4x1 + 2x16 = 36amu

Here is another take on this reaction:

Video from: Noel Pauller

• This key question hasn't been answered yet.

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