N_(2( g))+3 H_(2( g)) ⇌ 2 NH_(3( g))
20 g 5 g
Consider the above reaction, the limiting reagent of the reaction and number of moles of NH₃ formed respectively are :

Solution

Chemistry ma’am · 4 steps

Do not compare grams. Convert both to moles, then divide each by its coefficient in the equation.

1. Grams to moles. Molar masses are 28 g mol⁻¹ for N₂ and 2 g mol⁻¹ for H₂. So N₂ is 20/28 = 0.714 mol and H₂ is 5/2 = 2.5 mol.

   Reactant	Moles present	Coefficient	Moles ÷ coefficient
   N₂	0.714	1	0.714, smaller
   H₂	2.5	3	0.833

2. Limiting reagent. The smaller ratio belongs to N₂, so it runs out first. As a check, 0.714 mol of N₂ needs only 3 × 0.714 = 2.14 mol of H₂, and 2.5 mol is available.
3. Ammonia formed. Each mole of N₂ gives 2 mol of NH₃, so NH₃ = 2 × 20/28 = 10/7 ≈ 1.43 mol.
4. So. N₂ is the limiting reagent and about 1.43 mol of NH₃ forms. Option (c) lists this as 1.42 moles, so the answer is (c).

The idea to keep. To find the limiting reagent, convert every reactant to moles and divide by its coefficient. The smallest value runs out first, and it alone fixes how much product forms.

Answer from the NTA final answer key. Solution by PrepWiser.