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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 😀

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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