# How can I balance synthesis reactions?

May 10, 2014

A synthesis reaction is is the combining of two single elements into a compound. This can be accomplished either as an ionic or molecular coupling in the pattern

$A + B \to A B$

We will look at two examples of synthesis, first the formation of Ammonia $N {H}_{3}$ and the oxidation of iron to form rust.

To produce ammonia the basic equation is

$N + H \to N {H}_{3}$

We must remember that Nitrogen and Hydrogen are both diatomic molecules in their standard gas form. This adjusts the equation to

${N}_{2} + {H}_{2} \to N {H}_{3}$

Now we need to adjust coefficients in order to balance the atoms on each side of the equation. Currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

We can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

${N}_{2} + 3 {H}_{2} \to 2 N {H}_{3}$

This gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

The equation is balanced.

In the case of the formation of Rust $F {e}_{2} {O}_{3}$

The basic equation is

$F e + O \to F {e}_{2} {O}_{3}$

We are reminded that Oxygen is a diatomic and adjust the equation to

$F e + {O}_{2} \to F {e}_{2} {O}_{3}$

We now see 1 iron and 2 oxygen as reactants and 2 iron and 3 oxygen as products. We can balance the oxygen by placing a 2 coefficient in front of rust and a 3 coefficient in from of the oxygen to get 6 oxygen atoms on each side.

$F e + 3 {O}_{2} \to 2 F {e}_{2} {O}_{3}$

Since iron has only one atom on the reactant side we must use the coefficient 4 to balance the iron to 4 atoms on each side. To balance the final equation.

$4 F e + {O}_{2} \to F {e}_{2} {O}_{3}$