An sodium metal salt is prepared from #3.0*g# metal and #2.0*g# of iodine. What is the MAXIMUM possible mass of #NaI# that can be prepared?

1 Answer
anor277
Oct 26, 2016

Well, first you need a stoichiometric equation......

Explanation:

#Na + 1/2I_2 rarr NaI#

#"Moles of metal"# #=# #(3.0*g)/(22.99*g*mol^-1)# #=# #0.131*mol#.

#"Moles of halogen"# #=# #(2.0*g)/(2xx126.9*g*mol^-1)# #=# #1.24xx10^-4*mol#.

Now clearly, the metal is present in vast stoichiometric excess. At most, we can form #2.48xx10^-4*mol# of salt. I leave it to you to determine the mass of salt possible.

Note that the halogens, as elements, are all binuclear, #X_2#, and this is knowledge that the examiner expects. All elemental gases, save for the Noble Gases, and (maybe) mercury, are binuclear.

Related questions
See all questions in Percent Yield
Impact of this question
2004 views around the world
You can reuse this answer
Creative Commons License