What mass of dihydrogen gas results from the oxidation of #20*g# sodium metal by a mass of #10*g# of water?

1 Answer
anor277
May 29, 2017

Well first we write a stoichiometric equation, and get a bit over #1*g# of #H_2(g)#

Explanation:

We represent the reaction by the following stoichiometric equation:

#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#

You can double this if you like. I think the arithmetic is easier if you include the #1/2# coefficient.......

#"Moles of sodium"=(20*g)/(22.99*g*mol^-1)=0.870*mol#

#"Moles of water"=(10*g)/(18.01*g*mol^-1)=0.555*mol#.

Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only #0.555*mol# of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.

And thus we get #(0.555*mol)/2# dihydrogen gas evolved, i.e. #(0.555*mol)/2xx2.02*g*mol^-1=1.11*g#

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