1.89 g H2 is allowed to react with 9.91 g N2, producing 2.37 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

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thomasdecabooter thomasdecabooter · Answer 1 · 2020-03-13T06:25:13+01:00

Answer:

The theoretical mass of NH3 is 10.6 grams

Explanation:

Step 1: Data given

Mass of H2 = 1.89 grams

Mass of N2 = 9.91 grams

Mass of NH3 = 2.37 grams = actual yield

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

-Step 3: Calculate moles

Moles = mass / molar mass

Moles H2 = 1.89 grams / 2.02 g/mol

Moles H2 = 0.936 moles

Moles N2 = 9.91 grams / 28.0 g/mol

Moles N2 = 0.354 moles

Step 4: Calculate the limiting reactant

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

H2 is the limiting reactant. It will completely be consumed, 0.936 moles

N2 is in excess. There will react 0.936/ 3 = 0.312 moles N2

There will remain 0.354 - 0.312 = 0.042 moles

Step 5: Calculate moles NH3

For 3 moles H2 we'll have 2 moles NH3

For 0.936 moles H2 we'll have 2/3 * 0.936 = 0.624 moles NH3

Step 6: Calculate mass NH3

Mass NH3 = 0.624 moles * 17.03 g/mol

Mass NH3 = 10.6 grams

The theoretical mass of NH3 is 10.6 grams