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.

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