The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product. Consider the reaction: N2 (g) + 3 H2 (g) → 2 NH3 (g) If 13.90 g N2 is mixed with 6.440 g H2, calculate the theoretical yield (g) of NH3 produced by the reaction.

Question

Chemistry

Molly Walls

19 April, 09:03

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product. Consider the reaction: N2 (g) + 3 H2 (g) → 2 NH3 (g) If 13.90 g N2 is mixed with 6.440 g H2, calculate the theoretical yield (g) of NH3 produced by the reaction.

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Kayley Delgado · Answer 1

The theoretical yield of NH3 produced by the reaction is 16.89 grams

Explanation:

Step 1: Data given

Mass of N2 = 13.90 grams

Molar mass of N2 = 28.0 g/mol

Mass H2 = 6.440 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 13.90 grams / 28.0 g/mol

Moles N2 = 0.496 moles

Moles = 6.440 grams / 2.02 g/mol

Moles H2 = 3.19 moles

Step 4: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.496 moles)

H2 is in excess. There will react 3*0.496 = 1.488 moles

there will remain 3.19 - 1.488 = 1.702 moles

Step 5: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.496 moles N2 we will have 2*0.496 = 0.992 moles NH3

Step 6: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.992 moles * 17.03 g/mol

Mass NH3 = 16.89 grams

The theoretical yield of NH3 produced by the reaction is 16.89 grams