How many sodium atoms are there in 6.0 g of #Na_3N#?

1 Answer
anor277
Jun 29, 2016

#"Moles of sodium"# #=# #3xx"moles of sodium nitride"#

Explanation:

#"Moles of sodium nitride"# #=# #(6.0*g)/(82.99*g*mol^-1)#

Thus number of sodium atoms (or at least number of sodium ions):

#(6.0*g)/(82.99*g*mol^-1)xxN_Axx3#, where #N_A=6.022xx10^23*mol^-1#

i.e. #3xx(6.0*g)/(82.99*g*mol^-1)xx6.022xx10^23*mol^-1#. We get an actual number, without units, as required.

Related questions
See all questions in The Mole
Impact of this question
20400 views around the world
You can reuse this answer
Creative Commons License