In the reaction: $$\ce{3 Zn + 2 H3PO4 -> Zn3(PO4)2 + 3H2}$$

I know the initial mass of $\ce{Zn}$ to be 13.08 g and the volume of phosphoric acid to be 0.5 $\mathrm{dm^3}$. How do I work out the final concentration of $\ce{Zn(PO4)2}$ and the volume of $\ce{H2}$ gas evolved?

Once I work out the moles of Zn, are the moles of $\ce{Zn(PO4)2}$ just $\ce{Zn}$ moles/3?


closed as off-topic by ron, Klaus-Dieter Warzecha, jerepierre, Philipp, Jannis Andreska Mar 2 '15 at 18:04

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  • $\begingroup$ Your conclusion on the molar ratio of zinc phosphate and zinc is right and you'll figure out the rest too! $\endgroup$ – Klaus-Dieter Warzecha Mar 2 '15 at 16:10
  1. Your equation uses molar ratios, rather than weights or volumes.
  2. If you know the weight of zinc used, you know the number of moles. From this, you can calculate the number of moles of zinc phosphate generated.
  3. Supposed that the phosphoric acid is sufficiently concentrated to react all zinc, the number of moles calculated above together with the volume of the phosphoric acid used allows to calculate the concentration of zinc phosphate. Note: Do not omit units in your calculations!
  4. As far as the hydrogen is concerned: The number of moles produced can be calculated using the amount of zinc used. Additional knowledge of the molar volume of gases under standard conditions is helpful to figure out the volume.

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