Question

Ammonia is produced by the millions of tons annually for use as fertilizer. It is commonly made from N2 and H2 by the Haber process. Because the reaction reaches equilibrium before going completely to product, the stoichiometric amount of ammonia is not obtained. At a particular temperature and pressure, 12.17 g H2 reacts with 24.34 g N2 to form ammonia. When equilibrium is reached, 23.24 g NH3 has formed.

(a) Calculate the percent yield,
(b) How many moles of N2 and H2 are present at equilibrium?

Answer 1

The % yield is 78.15 %

At the equilibrium we have 3.418 moles of H2 and 0.869 moles of N2

Step-by-step explanation:

Step 1: Data given

Mass of H2 = 12.17 grams

Mass of N2 = 24.34 grams

Mass of NH3 formed = 23.24 grams

Molar mass H2 = 2.02 g/mol

Molar mass N2 = 28 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of H2

Moles H2 = Mass H2 / molar mass H2

Moles H2 = 12.17 g / 2.02 g/mol

Moles H2 = 6.025 moles

Step 4: Calculate moles of N2

Moles N2 = 24.34 g / 28.0 g/mol

Moles N2 = 0.869 moles

Step 5: Calculate limiting reactant

For 1 mole of N2 consumed, we need 3 moles of H2 to produce 2 moles of NH3

N2 is the limiting reactant. It ill completely be consumed. (0.869 moles)

H2 is in excess. There will react 3*0.869 = 2.607 moles

There will remain 6.025 - 2.607 = 3.418 moles of H2

Step 6: Calculate moles of NH3

For 1 mole of N2 consumed, we need 3 moles of H2 to produce 2 moles of NH3

For 0.869 moles of N2 we have 2*0.869 = 1.738 moles of NH3

Step 7: Calculate mass of NH3

mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 1.738 moles * 17.03 g/mol

Mass NH3 = 29.60 grams NH3 = theoretical yield

Step 8: Calculate % yield

%yield =(actual yield / theoretical yield)*100%

% yield = (23.24 grams / 29.60 grams) * 100%

% yield = 78.51 %

The % yield is 78.15 %

At the equilibrium we have 3.418 moles of H2 and 0.869 moles of N2