Mg3 N3 (s) + 6H20(1)---3Mg(OH)2 + 2NH3 (g)?

A

askifitsreal

New member
Mg3 N3 (s) + 6H20(1)---3Mg(OH)2 + 2NH3 (g)?

When 100g of Mg3N2 reacts with 75.0g of H2O, what is the limiting reactant?
1.56mol Mg3N2 is the limiting agent

When 100g of Mg3N2 reacts with 75.0g H2O, what is the maximum theoretical yield of NH3?
8.84gNH3

When 100g of Mg3N2 reacts with 75 g H2O, 15.0 g NH3 is formed. What is the % yield?

I just need to figure out this one....
 
R

RogertheMole

New member
Mg3N2 + 6 H2O -> 3 Mg(OH)2 + 2 NH3

(100 g Mg3N2) / (100.9286 g Mg3N2/mol) = 0.990799 mol Mg3N2
(75.0 g H2O) / (18.0153 g H2O/mol) = 4.16313 mol H2O
4.16313 mol H2O would react completely with 4.16313 x (1/6) = 0.693855 mol Mg3N2, but there is more Mg3N2 present than that, so Mg3N2 is in excess and H2O is the limiting reactant.

(4.16313 mol H2O) x (2/6) x (17.0306 g NH3/mol) = 23.6 g NH3

(15.0 g NH3) / (23.6 g NH3) = 0.636 = 63.6%
 
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