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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jaxtothemax on October 24, 2016, 12:09:42 PM

Title: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: jaxtothemax on October 24, 2016, 12:09:42 PM
Hello chemists!

I need help with an exercise we got at school. I am losing my mind and I'm nervous as hell because I can't figure it out and I bet it's easy as hell...

So we got this formula:

CaH2 + 2H2O -> Ca(OH)2 + 2H2

and we need to calculate how many grams of Ca(OH)2 we get after the reaction. The only info we have is that there is 15g of CaH2. Also it says that water is negligible, so we count that out. But I'm kind of confused because of the last statement (that water is negligible), because I don't know what to count on the right side of the chemical equation. One atom of oxygen and 2 atoms of hydrogen don't count on the other side? Say what?  Anyway, could someone please enlighten me.

Thanks  :)

Best regards!
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: Arkcon on October 24, 2016, 12:32:47 PM
Hard to know what the question means by "negligible" and what you mean as well. 

You realize that the equation has to be balanced -- same atoms either side.  It's not negligible to have that, and you do, anyway.  So that's not a problem.

They've given you how much of one reactant, and want to know how much product.  I'm guessing they mean you can have all the water you want, so limitations on that amount are negligible.
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: Borek on October 24, 2016, 02:14:37 PM
Like Arkcon said, hard to guess what "negligible" is intended to mean in this context. Probably that you can ignore it in the calculations, assuming there is enough.

We typically describe such situation in stoichiometry problems by saying something like "15 g of CaH2 reacted with excess of water".
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: jaxtothemax on October 24, 2016, 02:30:18 PM
Yes, it means that you can ignore it. I'm not a native English speaker, especially when it comes terminology... Sorry about that. So any clue how to calculate that?
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: AWK on October 24, 2016, 05:09:22 PM
You can calculate amount of Ca(OH)2 or H2 (as mass or volume) or both.
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: Borek on October 24, 2016, 06:07:21 PM
This is just a simple stoichiometry - balance the reaction (you can omit this step if the reaction is already balanced), convert mass to moles, use the stoichiometric ratio to calculate number of moles of the product, convert to mass.
Title: Re: CaH2 + 2H2O -> Ca(OH)2 + 2H2 (15g CaH2, water negligible, how much Ca(OH)2)
Post by: jaxtothemax on October 25, 2016, 11:53:57 AM
Now that I know this kind of stuff is named "Stoichiometry" in English I did a research and found some explanations. So I got 0.36mol of Ca(OH)2 and 0.72mol of H2. It's just all a matter of the numbers in front of the elements/compounds. Thanks guys!