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

1 Answer
Jan 31, 2016

Answer:

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.