# Writing Chemical equation given molarity

If I am given the molarity of two substances and I know how they react, how can I write the balanced chemical reaction? I.e.,

Experimentally I found:

1.0 M $\ce{NH_3(aq)}$ mixed with 0.1 M $\ce{BaCl_2(aq)}$ yielded no precipitate when 1ml of each were mixed together. My unbalanced chemical equation so far looks like

$$\ce{N^{-3}H^+_3(aq) + (1/10)Ba^{+2}Cl^{-}_2(aq) -> N^{-3}Ba^{+2} + Cl^{-2}H^+}$$

My logic being: since the solutions are administered in equal volumes, their molarity determines the molar coefficients. Right off the bat I can tell that $\ce{NH}$ is not soluble, but the other compound I think is, which would contradict my observations. Either way, I am incredibly lost. Any help greatly appreciated.

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## 1 Answer

Have you considered the possibility that your reaction is wrong. where, for example, is the barium on the product side. I'll give you one hint here: barium chloride dissociates in water to barium and chloride ions. Ammonia reacts with water to produce some ammonium ion and the same amount of hydroxide ion. However, barium hydroxide is totally ionized.

Can you take it from there?

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I've fixed the mistake, but I'm still not sure what to do. Do I use solubility rules now? Should I even bother balancing the chemical equation? –  Cactus BAMF Nov 4 '12 at 16:17
Hint: List the possible products and determine which ones are insoluble. –  Paul J. Gans Nov 5 '12 at 18:30