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

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Answer 1 · 2016-06-29T07:47:59

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

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

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.

How many sodium atoms are there in 6.0 g of Na_3NNa_3N?