Gen chem question

[paragon1111]

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hi everyone,

This should be a simple question, but I can't seem to reason out where i went wrong.

When Sodium reacts with excess water, the products are

A) Na2O
B) Na2O + NaOH
C) Na2O + NaOH + H2
D) NaOH + H2
E) NaOH

I'm wondering why Na2O cannot be a product. Since NaOH is a strong base, doesn't it dissociate completely, thus pushing equilibrium back in the other direction?

Thanks

 

[shallsen]

Check out this thread, same question
http://forums.studentdoctor.net/threads/sodium-reaction-with-excess-water-gen-chem.1015015/

 

[DAT Destroyer]

hi everyone,

This should be a simple question, but I can't seem to reason out where i went wrong.

When Sodium reacts with excess water, the products are

A) Na2O
B) Na2O + NaOH
C) Na2O + NaOH + H2
D) NaOH + H2
E) NaOH

I'm wondering why Na2O cannot be a product. Since NaOH is a strong base, doesn't it dissociate completely, thus pushing equilibrium back in the other direction?

Thanks

No....This is NOT an equilibrium-controlled process. It turns out that having a base produced and a gas is energetically more favorable than the oxide. This is an important reaction that you should remember.

Hope this helps.

Dr. Romano

 

[Denevistas]

I just reasoned this out as Na is a group 1 metal that wants to get oxidised. It will thus favor an ionic bond.

 

[Denevistas]

I believe this is more about knowing conversions, but I am unable to work around it...help?

 

[Frank22]

Think of the reaction this way:

Na + HOH

Because Na becomes a cation, it will bond to the OH of the water, and producing H2, so the reaction is as such:

NaOH + H2

Remember that alkali metals react extremely violently with water; the same applies to the alkaline earth metals, but to a lesser extent.

 

[paragon1111]

Thanks for all your helpful replies everyone!

 

[Denevistas]

Need

help on this one!

 

[Lnguyen7]

NeedView attachment 197559 help on this one!

I'll try this one.

q = msΔT

Divide both sides by t (time)
q/t = msΔT/t (1)

D = m/V, so m = DV
Substitute m into (1): q/t = DVsΔT/t = D(V/t)sΔT (2)

Units q (J), t (s), D (kg/m^3), V (m^3), t (s), s [J/(gC)], ΔT (C)

W is the unit of power and 1 W = 1 J/s

A = 70,000 m^2 and rate of solar energy incident is 120 W/m^2 or 120 J/(m^2*s)
So the total power needed is 70,000*120 = 8,400,000 (J/s)

You want to find (V/t) with V in L
(2): 8,400,000 (J/s) = (1,000 kg/m^3)(V/t)[4.2 J/(gC)](10 C)(1,000 g/1 kg)
(V/t) = 0.2 m^3/s = 200 L/s

 

[Denevistas]

Thanks!