NEET + JEEAdvanced

In the Haber process, N2 + 3H2 -> 2NH3. A reaction vessel is charged…

Question

In the Haber process, N2 + 3H2 -> 2NH3. A reaction vessel is charged with 28.0 g of N2 (molar mass 28) and 8.0 g of H2 (molar mass 2). After reaction, 25.5 g of NH3 (molar mass 17) is actually obtained. Identify the limiting reagent and calculate the percentage yield of ammonia.

✓ Verified answer: 75checked by our engine — not a guess

Step-by-step solution

Moles N2 = 28.0/28 = 1.00; moles H2 = 8.0/2 = 4.00.

N2 needs 3 mol H2 per mol; 1.00 mol N2 needs 3.00 mol H2, and 4.00 mol H2 is available, so N2 is the limiting reagent.

Theoretical NH3 = 2 x 1.00 = 2.00 mol = 34.0 g.
%yield = 25.5/34.0 x 100 = 75.0%.

Final answer75.0

Stuck on a problem like this?

Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.

Solve my doubt →

More Phys Chem Mole solutions