When heated, lithium reacts with nitrogen to form lithium nitride: 6li(s) + n2(g) → 2li3n(s) what is the theoretical yield of li3n in grams when 12.7 g of li is heated with 34.7 g of n2? g if the actual yield of li3n is 5.85 g, what is the percent yield of the reaction? %

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 %

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because it does not have the potential to be entirely used up

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