Question

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

Answer 1

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`