Limiting reagent
56.3 kg NH₃H₂ is the limiting reagent
Molar mass
Molar mass. The mass of one mole in grams: add the atomic mass of every atom in the formula. Masses. Standard atomic masses to two decimals, hydrogen 1.008, every element to uranium. Typing. Ca(OH)2, (NH4)2SO4, CuSO4.5H2O and K4[Fe(CN)6] all work.
18.02g/mol
H₂O
- H× 211.19%
- O× 188.81%
- H2 × 1.0082.016
- O1 × 16.0016.00
- Total18.02 g/mol
Grams, moles, particles
Type any one. The other three fill in. Moles. From grams, particles or litres they are rounded to 3 significant figures first, as in a worked answer. 1 mol = 6.022 × 10²³ particles = 22.7 L of gas at STP (1 bar); 22.4 L at 1 atm.
0.5 mol of H₂O = 3.011 × 10²³ molecules
At 1 atm (older value): 11.20 L
Empirical formula
1. Take 100 g, so each percentage is grams. 2. Divide by the atomic mass to get moles, then by the smallest. 3. Multiply by 2, 3 or 4 if a ratio ends in .5, .33 or .25. Molecular formula. Add the molar mass: n = molar mass ÷ empirical formula mass.
%
%
%
Empirical formulaCH₂Cl49.48 g/mol
Molecular formulaC₂H₄Cl₂n = 98.96 ÷ 49.48 = 2
| Element | g in 100 g | ÷ mass | ÷ least | Whole |
|---|---|---|---|---|
| H | 4.07 | 4.038 | 2.00 | 2 |
| C | 24.27 | 2.021 | 1.00 | 1 |
| Cl | 71.65 | 2.021 | 1.00 | 1 |
Limiting reagent
1. The equation is balanced for you. 2. Each mass becomes moles, divided by its coefficient. 3. The smallest runs out first: it is the limiting reagent and sets how much product forms. Leave a mass empty if that reactant is in excess.
N₂ + 3 H₂ → 2 NH₃
- N₂1784 mol ÷ 1 = 17843.67 kg left over
- H₂4960 mol ÷ 3 = 1653Limiting
Formed
- NH₃3307 mol56.3 kg
The NCERT way
moles ÷ coefficient for each reactant: the smallest is the limiting reagent
50.0 kg of N₂ and 10.0 kg of H₂ are mixed to make ammonia. Which is the limiting reagent, and how much NH₃ forms?
- Balanced equation: N₂ + 3 H₂ → 2 NH₃.
- Moles of N₂ = 50,000 g ÷ 28.02 = 1784 mol. Moles of H₂ = 10,000 g ÷ 2.02 = 4960 mol.
- 1784 mol of N₂ needs 3 × 1784 = 5353 mol of H₂, but only 4960 mol is there. So H₂ is the limiting reagent.
- NH₃ formed = 4960 × 2/3 = 3307 mol = 3307 × 17.03 g = 56.3 kg.
- H₂ is the limiting reagent and 56.3 kg of NH₃ forms.
Limiting reagent: quick answers
What is a limiting reagent?
It is the reactant that is used up first. Once it runs out the reaction stops, so it decides how much product forms. The other reactants are left over in excess.
How do you find the limiting reagent?
Turn each mass into moles and divide by the coefficient of that reactant in the balanced equation. The smallest answer belongs to the limiting reagent.
How much ammonia forms from 50.0 kg N₂ and 10.0 kg H₂?
H₂ is limiting, so 4960 mol of H₂ gives 3307 mol of NH₃, which is 56.3 kg.