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

Question

Ellie Andrews · Accepted 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.

#Na(s) + H_2O(g) rarr 1/2H_2(g) + NaOH(aq)# With this stoichiometry, half a mole of dihydrogen evolves for every mole of sodium. 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.

Serenity Allen · Answer

undefined To calculate the volume of hydrogen generated from 2 grams of sodium, you can use the stoichiometry of the reaction between sodium and water:

2 Na + 2 H2O -> 2 NaOH + H2

Using the molar mass of sodium (Na) and the molar volume of hydrogen gas at standard temperature and pressure (STP), you can determine the volume of hydrogen produced.