2013年7月28日 · The reaction of sodium with water is quite infamous. $\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$ But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$?

2015年5月16日 · In the complex ion $\\ce{[Na(H2O)6]+}$, the sodium cation forms 6 coordinate bonds with water ligands. Typically this octahedral form is associated with $\\mathrm{sp^3d^2}$ hybridization as far as I ...

2017年1月26日 · $$\ce{2 Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$ In the typical setup, you have lots of water and little sodium. Therefore, whatever you get will be a solution or a precipitated salt. Alkali metals, however, often form very soluble salts and indeed all alkali hydroxides are well soluble.

2015年11月27日 · How would you balance: Na + H2O --> NaOH + H2? Chemistry Chemical Reactions Balancing Chemical Equations ...

2016年1月3日 · Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr Sodium is oxidized; hydrogen is reduced. You can double the equation if you like to remove the 1/2 coefficient, but the equation is stoichiometrically balanced.

2017年9月11日 · 2Na + 2H_2O -> 2NaOH + 2H_2 First, let's rewrite H_2O as HOH. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the "new" equation is: Na + HOH -> NaOH + H_2 Now we can see how many of each element/polyatomic ion we have on each side of the equation. Left: Na - 1 H - 1 OH - 1 Right: Na - 1 OH -1 H -2 Looking at what we have, there is an unequal amount ...

2015年2月5日 · $$ \begin{align} \ce{NaOH + H2O &-> Na+ + OH- + H2O}\\ \ce{K + H2O + OH- &-> K+ + 2OH- + H2} \end{align} $$ Essentially, the potassium is going to react with the water long before anything else. The only way I can think of to get your reaction to go forward would be to have the molten forms of $\ce{NaOH}$ and $\ce{K}$, which would react like this:

2017年5月25日 · For the reaction #2Na + 2H_2O -> NaOH + H_2#, how many grams of hydrogen are produced from 130 g of sodium?

2018年1月29日 · Is this not an acid-base reaction? underbrace(NaH(s))_"the base" + underbrace(H_2O(l))_"the acid" rarr NaOH(aq) + H_2(g)uarr

2015年8月30日 · $$\ce{(something)OH + H+ -> (something)+ + H2O}$$ would be water, and that seems unsettling to me. I also believe that since $\ce{NaOH}$ undergoes the following reaction: $$\ce{NaOH -> Na+ + OH-}$$ the $\ce{Na+}$ is something of a 'spectator ion' (not sure if that's the correct term), this seems to imply that $\ce{H2O}$ should be the conjugate ...