# Question cbc36

May 7, 2015

All you need to do to solve this problem is look at the mole ratios that exist between the species that take part in the reaction.

$\textcolor{red}{2} A {l}_{2} {O}_{3} \to \textcolor{b l u e}{4} A l + \textcolor{g r e e n}{3} {O}_{2}$

For a balanced chemical equation, the stoichiometric coefficients will tell you the relationship that exists between the number of moles of each species.

In your case, $\textcolor{red}{2}$ moles of aluminium oxide will produce $\textcolor{b l u e}{4}$ moles of aluminium metal and $\textcolor{g r e e n}{3}$ moles of oxygen gas. These mole ratios will always be true for the number of moles that react, regardless of how many moles of each species you're dealing with.

So, if 6 moles of aluminium oxide, you'll produce

6cancel("moles"Al_2O_3) * (color(green)(3)"moles"O_2)/(color(red)(2)cancel("moles"Al_2O_3)) = "9 moles" ${O}_{2}$

Likewise, if 1 mole of aluminium metal was produced by the reaction, it must mean that

1cancel("mole Al") * (color(red)(2)"moles"Al_2O_3)/(color(blue)(4)cancel("moles Al")) = "0.5 moles" $A {l}_{2} {O}_{3}$

In similar fashion, if 4 moles of oxygen gas are produced, you'll also get

4cancel("moles"O_2) * (color(blue)(4)"moles Al")/(color(green)(3)cancel("moles"O_2)) = "5.33 moles Al"#

As a conclusion, always be aware of the mole ratios that exist between the species, as they are great tools for stoichiometric problems.