38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. how many grams of silicon nitride can be formed if nitrogen is the limiting reactant?
3si + 2n2 yields si3n4

3Si + 2N2 --> Si3N4 (as given) 

n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol

Therefore, N2 is limiting and Si is in excess 
The molar ratio of 2N2:Si3N4 is 2:1 
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2) 

m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g 
= 8.066g (4 significant figures) 

(hopefully it is right, but double check in case i did something wrong) :)