A beaker has 0.2 M of Na2SO4. What will be the concentration of sodium and sulfate ions?

Na+-0.0648M

SO4²-0.1352M

Explanation:

First find the R.A.M of the sodium sulphate

(23×2)+(32)+(16×4)=142

Find the portion of sodium ions=46/142×0.2

=0.0648M

Sulphate=

0.2-0.0648=0.1352M

Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g