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








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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 😀




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+. 


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