# How can I do stoichiometry mole to mole problems?

Oct 27, 2015

By learning how to balance chemical equations.

#### Explanation:

Let's take a look at the simplest example:

$2 {H}_{2}$ + ${O}_{2}$ = $2 {H}_{2} O$ (balanced equation)

Given the equation above, how many moles of ${O}_{2}$ will react with sufficient number of moles of ${H}_{2}$ do you need to produce 10 moles of ${H}_{2} O$?

From the balanced equation, already know that for every 1 mole of ${O}_{2}$, it produces 2 moles of ${H}_{2} O$, or simply put:

1 mole ${O}_{2}$ = 2 moles ${H}_{2} O$

Therefore, in order to get the number of moles for ${O}_{2}$,

$10$ $\cancel{\text{mol}}$ ${H}_{2} O$ x (1 "mol" O_2)/(2 cancel ("mol") H_2O) = $5$ $\text{mol}$ ${O}_{2}$

If 5 moles of ${O}_{2}$ are needed to produce 10 moles of ${H}_{2} O$, how many moles of ${H}_{2}$ do you need to react?

Again, from the balance equation, you will see that every 1 mole of ${O}_{2}$ needed 2 moles of ${H}_{2}$ to produce 2 moles of ${H}_{2} O$, or simply put,

1 mole ${O}_{2}$ = 2 moles ${H}_{2}$

Thus,

$5$ $\cancel{\text{mol}}$ ${O}_{2}$ x (2 "mol" H_2)/(1 cancel ("mol") O_2) = $10$ $\text{mol}$ ${H}_{2}$

See? Everything starts by properly balancing the equation first because that is where you can get your "conversion factors" in order to solve the mole-to-mole problems.