4. If 100.0g of nitrogen gas (N2) is reacted with 100.0g of hydrogen gas (H2) to form NH3. What are the limiting and excess reactants?

Hint: Convert grams to moles for each reactant and then convert to moles of NH3. You need your balanced equation from answer 1 to determine the mole relationship between each reactant and the product NH3. Use the periodic table to determine the molar mass of all chemical formulas. Fill in the “?” blanks below to show your work.

Balanced equation: 1N2 (g) + 3H2(g)  2NH3(g)

Molar mass of H2 = g/mol

100.0"g " "H" _2×(1"mol " "H" _2)/( ? "g " "H" _2 )×(" ? mol N" "H" _3)/(" ? mol " "H" _2 )= ? "mol N" "H" _3

Molar Mass of N2 = g/mol

100.0"g " "N" _2×(1"mol " "N" _2)/(" ? g " "N" _2 )×(" ? mole N" "H" _3)/(" ? mol " "N" _2 )= ? "mol N" "H" _3

A. What is the excess reactant? Hint: it is the reactant above that makes the most amount of product

B. What is the limiting reactant? Hint: it is the reactant above that makes the least amount of product

C. What is the theoretical yield of ammonia of the reaction?
Molar mass of NH3 = g/mol

100.0"g " "N" _2×(1"mol " "N" _2)/( ? "g " "N" _2 )×( ? "mol N" "H" _3)/( ? "mol " "N" _2 )×(" ? g N" "H" _3)/( 1 "mol N" "H" _3 )= ? "g N" "H" _3

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LammettHash LammettHash · Answer 1 · 2020-12-01T22:22:27+01:00

Nitrogen (N) has a molar mass of about 14.007 g/mol, so you have

100.0 g N₂ = (100.0 g) • (1/28.014 mol/g) ≈ 3.570 mol N₂

Hydrogen (H) has a molar mass of about 1.008 g/mol, so

100.0 g H₂ = (100.0 g) • (1/2.016 mol/g) ≈ 49.60 mol H₂

In the balanced reaction, 1 mole of N₂ reacts with 3 moles of H₂ to make 2 moles of NH₃. We have

3.570 • 3 ≈ 10.71

so the reaction would use up all 3.570 mol N₂ and 10.71 mol H₂ to produce about 7.139 mol NH₃.

(A) H₂ is the excess reactant. There is an excess of 49.60 - 10.71 ≈ 38.89 mol H₂.

(B) N₂ is the limiting reactant. All of the N₂ gets consumed in the reaction.

(C) NH₃ has a molar mass of (14.007 + 3 • 1.008) g/mol = 17.031 g/mol. This reaction would theoretically yield about 7.139 mol NH₃, or

7.139 mol NH₃ = (7.139 mol) • (17.031 g/mol) ≈ 121.6 g NH₃