If 74.0 sodium is mixed with 27.0 nitrogen gas, what mass of sodium nitride forms?

Question

If 74.0  sodium is mixed with 27.0  nitrogen gas, what mass of sodium nitride forms?

anonymous · Answer

6Na + N2 = 2Na3N        Na will be the limiting factor. 

M.W of NA = 23 74/23 = 3.22 moles 3.218 /6 = .536 moles 
M.W of N = 14 27/14 = 1.93 moles  /2 = .964 moles

 3.22 moles Na * (2 moles Na3N) / 6 Moles of NA = 1.07 moles of Na3N formed. 
1.07 moles Na3N * 83 (M.W of Na3N) = 88.71 grams of Na3N formed. 
WOW that took forever :(

anonymous · Answer

Thank You some much :D

anonymous · Answer

If the reaction in  has a percent yield of 69.0 %, how much sodium nitride is actually produced?