# My question is about stoichiometry, limiting reagent problems [closed]

So I'm really struggling with this particular concept of these problems. I know how to convert to moler mass and all that, but I don't know how to do the ratio part, I can do them when the ratio is 1/2 or 1/3 but I don't know what to do with it when the ratio is 5/3 or others that don't include a 1. Any help would be greatly appreciated, thanks 😀

## closed as unclear what you're asking by Tyberius, andselisk♦, Zhe, Mithoron, airhuffOct 13 '17 at 22:45

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• What's the difference? Presumably, you are multiplying with the ratio. Why does it matter if there is a 1? – Zhe Oct 13 '17 at 21:43

Let's look at the reaction between permanganate and oxalate in acid as an example.

$\ce{5C2O4^2- + 2MnO4^- + 16H^+(aq) -> 2Mn^2+ + 10CO2(g) + 8H2O}$

So if I have 13 moles of oxalate and 6 moles of permanganate what is the limiting reagent?

An easy way to solve is to divide the chemical equation by 2 to get

$\dfrac{5}{2}\ce{C2O4^2- + MnO4^- + 8H^+(aq) -> Mn^2+ + 5CO2(g) + 4H2O}$

So $\dfrac{\text{moles oxalate}}{\text{moles permanganate}} = \dfrac{5}{2} = 2.5$

Given that we started with

$\dfrac{\text{moles oxalate}}{\text{moles permanganate}} = \dfrac{13}{6} = 2.167$

there is too little oxalate to react with all the permanganate. For a complete reaction we'd need 6 moles of permanganate and $\dfrac{5}{2}(6) = 15$ moles of oxalate.

Also note as a different problem we can look at oxalate and $\ce{H+}$.

$\dfrac{\text{moles } \ce{H+} }{\text{moles oxalate}} = \dfrac{16}{5} = 3.2$

So if we started with 13 moles of oxalate, we'd need (3.2)(13) = 41.6 moles of H+.