Answer:
The % yield of the reaction = 27.5 %
Explanation:
Step 1: Data given
Mass of Li = 12.7 grams
Mass of N2 = 34.7 grams
Actual yield of Li3N = 5.85 grams
Molar mass of Lithium = 6.94 g/mol
Molar mass of N2 = 28 g/mol
Molar mass of LI3N = 34.83 g/mol
Step 2: The balanced equation:
6Li(s) + N2(g) → 2Li3N(s)
Step 3: Calculate moles of Lithium
Moles Li = mass Li / Molar mass Li
Moles Li = 12.7 grams / 6.94 g/mol
Moles Li = 1.83 moles
Step 4: Calculate moles of N2
Moles N2 = 34.7 g/ 28 g/mol
Moles N2 = 1.24 moles
Step 5: Limiting reactant
For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N
Lithium is the limiting reactant. It will completely be consumed (1.83 moles).
N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles
There will remain 1.24 - 0.305 = 0.935 moles
Step 6: Calculate moles of Li3N
For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N
For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N
Step 7: Calculate mass of Li3N
Mass Li3N =moles LI3N * Molar Mass LI3N
Mass Li3N = 0.610 moles * 34.83 g/mol
Mass Li3N = 21.2463 grams = Theoretical yield
Step 8: Calculate % yield
% yield = actual yield / theoretical yield
% yield = (5.85 / 21.2463)*100% = 27.5%
The % yield of the reaction = 27.5 %
(atoms of C in one mole) = 6.84*