Ambiguous reaction question. Help?

[Swenis]

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Hey guys and gals, this practice problem I started says that "a piece of sodium metal undergoes complete reaction with water. The gaseous product is collected over water at 25.5oC in a 246mL container at 745 torr. Provide a balanced equation for the reaction." Is this sodium metal just Na(s) or Na2CO3(aq) or what?

 

[DHG]

Hey guys and gals, this practice problem I started says that "a piece of sodium metal undergoes complete reaction with water. The gaseous product is collected over water at 25.5oC in a 246mL container at 745 torr. Provide a balanced equation for the reaction." Is this sodium metal just Na(s) or Na2CO3(aq) or what?

Na metal (just pure Na solid) reacts with water to form hydrogen gas (g) and sodium hydroxide (aq)...

2Na (s) + 2H2O (l) ----> H2(g) + 2NaOH (aq)

hope that hleps.

 

[Swenis]

Na metal (just pure Na solid) reacts with water to form hydrogen gas (g) and sodium hydroxide (aq)...

2Na (s) + 2H2O (l) ----> H2(g) + 2NaOH (aq)

hope that hleps.

Thank you, I just figured it out right before I refreshed this post 🙂

How do I find the concentration of NaOH if the Na(s) was originally added to 20.0 mL of water?

 

[DHG]

Thank you, I just figured it out right before I refreshed this post 🙂

How do I find the concentration of NaOH if the Na(s) was originally added to 20.0 mL of water?

use PV = nRT from the given pressure and volume info to solve for n moles of H2 gas. You know it takes 2 moles of Na for every 1 mole of H2 gas produced (from the balanced equation) so multiply moles H2 by 2. Now you know how many moles Na you have, which is = to the moles NaOH produced. You divide this (moles Na = moles NaOH) by how many L water you had, that will give you molarity NaOH.

 

[Swenis]

The only question I have is about the amount of volume (L) I need to use for determining molarity since the problem says that I need to determine the conc. of the product if the sodium was added to 20mL of water. So does the 246mL container have any relevance to this problem or do I just use .02L (20mL)? Thanks DHG, you're awesome. 👍

 

[DHG]

The only question I have is about the amount of volume (L) I need to use for determining molarity since the problem says that I need to determine the conc. of the product if the sodium was added to 20mL of water. So does the 246mL container have any relevance to this problem or do I just use .02L (20mL)? Thanks DHG, you're awesome. 👍

The way I understood the question, if it's written word for word in the original post, was that the 246mL container was the collection container for the H2 gas, so that is the V you would use in PV=nRT for the gas part...

P=745 torr
V= 246mL
T=25C
R=whichever one you want to use, constant
n= unknown, moles of H2 gas produced

Once you get n moles from this, multiply it by 2 to get moles of Na (and therefore moles NaOH) ...divide this # by 20mL water (0.02L) to get molarity of NaOH solution.

 

[Swenis]

Thanks, man. Yeah, it was written word for word. I appreciate it!

 

[vafcarrot]

Hey guys and gals, this practice problem I started says that "a piece of sodium metal undergoes complete reaction with water. The gaseous product is collected over water at 25.5oC in a 246mL container at 745 torr. Provide a balanced equation for the reaction." Is this sodium metal just Na(s) or Na2CO3(aq) or what?

Um, doesn't solid Na cause a huge explosion when exposed with air or water? I thought that was why it was stored under mineral oil or something.

 

[Swenis]

Um, doesn't solid Na cause a huge explosion when exposed with air or water? I thought that was why it was stored under mineral oil or something.

I don't think my professor expects my class' simple minds to figure that out. 😀