Lithium and nitrogen react in a combination reaction to produce
lithium nitride:
6Li(s) + N2(g) → 2Li3N(s)
In a particular experiment, 3.00 g samples of each reagent are
reacted. The theoretical yield of lithium nitride is ________
g.
6 Li(s) +N2(g) 2Li3N(s)
number of moles of Li= amount of Li (g)/ molar mass of Li= 3 g/ 6.94 g/mol= 0.4323 moles
number of moles of N2= amount of N2 (g)/ molar mass of N2= 3 g/ 28.01 g/mol= 0.1071 moles
According to balance equation
6 moles of Li requires 1 mole of N2
therefore 0.4323 moles of Li require will require= 1/6 X 0.4323 mole of N2= 0.0721 moles of N2
hence N2 is excess reagent
According to balance equation
6 moles of Li produces 2 moles of Li3N
therefore
0.4323 moles of Li will produces= 2/6 X 0.4323 moles of Li3N = 0.1441 moles of Li3N
Amount of Li3N produced in g= number of moles of Li3N produced X mw of Li3N (g/mol)
= 0.1441 mole X 34.82 g/mole = 5.02 g answer
Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) + N2(g) →...
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