Question

How to calculate the volume of hydrogen which generates from 2g of sodium?

Answer 1

We specify room temperature, atmospheric pressure, and predict a volume of approx `2` `dm^3` `H_2(g)`.

We also ASSUME that the `2.0` `g` of sodium metal has been dropped into a bucket of water, and the gas evolved is collected by displacement of water in a separate graduated container.

Explanation:

`Na(s) + H_2O(g) rarr 1/2H_2(g) + NaOH(aq)`

Given this stoichiometry, for every mole of sodium, one half a mole of dihydrogen is evolved.

Moles of `Na` `=` `(2.0*cancel(g) )/(22.99*cancel(g)*mol^-1)` `=` `0.0870` `mol`.

Given the stoichiometry, `0.0435` `mol` dihydrogen gas are evolved.

Now it is a fact, that at room temperature (`298` `K`) and `1` `atm`, 1 mole of ideal gas has a volume of `25.4` `dm^3`. (These molar volumes are often given in a examination!)

So given the conditions stated and assuming ideality, `25.4` `dm^3*mol^-1` `xx` `0.0870*mol` `=` `2.2` `dm^3` `=` `2.2` `L` volume of dihydrogen gas.