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I'm wondering if it's possible to balance a chemical equation with an incorrect mole ratio. I have the unbalanced equation $\ce{AgNO3 + Cu -> Ag + Cu(NO3)2}$, I have a $2:3$ ratio of $\ce{Cu}$ and $\ce{Ag}$ that I must plug in. Is this possible, because it doesn't look like it.

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The balanced reaction is $$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$

Firstly, it's not necessary that all the reactants are depleted in the reaction.

There is the concept of a "limiting reagent" -- the reagent that gets depleted first and drives the extent of the reaction.

In this case, the limiting reagent is $\ce{AgNO3}$, and you will have half a unit of $\ce{Cu}$ left over.

In the general case, the limiting reagent need not be the one in the least quantity. For example, for $\ce{2A +B ->C +D}$ , if $\ce{A:B}$ is 3:2, then $\ce{A}$ is still the limiting reagent as it will get depleted first.

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  • $\begingroup$ But I have to write an equation with the mole ratio of 2:3 $\endgroup$ – Someone Apr 7 '13 at 3:36
  • $\begingroup$ @Someone: Not possible unless you modify the reaction. $\endgroup$ – ManishEarth Apr 7 '13 at 3:43
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The balanced reaction is this: $$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$

Which is normally what you would write and then just mention that your reactants ratio is different so you have a limiting species. However, if you really want the 3:2 ratio inside your the reaction equation you could write this:

$$\ce{3AgNO3 + 2Cu -> 3 Ag + 3/2 Cu(NO3)2 + Cu}$$

where I have just added equal amounts of Cu and Ag on the left and right and explicitly put in the copper that is left over. Of course by standard convention (leaving no fractions in the reaction equation) you would actually have to write:

$$\ce{6AgNO3 + 4Cu -> 6 Ag + 3 Cu(NO3)2 + 2Cu}$$

So now you have a reactants ratio $\ce{Ag:Cu}=3:2$ and you can see that this will result in left over copper, because the silvernitrate is the limiting species.

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  • $\begingroup$ But I can't add 2 Cu to the other side. $\endgroup$ – Someone Apr 7 '13 at 14:58
  • $\begingroup$ @Someone : why not? if you put in 6 AgNo3 and 4 Cu that is what comes out: 2 Cu left over $\endgroup$ – Michiel Apr 7 '13 at 15:07

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