1) You need the balanced equation of the reaction.
2) Compute for molar mass of each reagent.
3) Kunin mo yung equivalent moles nila using the given amounts.
4) Multiply each of their moles to the ratio of mole of product (choose any) over mole of reagent.
5) Compare the resulting moles. Kung ano yung may mas mababang result, yun ang limiting reagent.
For example, 4NH₃ + 5O₂ ––> 4NO + 6H₂O (step 1: balanced na yan)
Given: 2.5g NH₃ and 2.85g O₂
Step 2: (use periodic table, kung di mo kabisado mass nila)
1 mole NH₃ =
17.03 g
1 mole O₂ = 32.0 g
Step 3 and 4: (i chose NO for the ratio)
[tex]2.5g NH_{3} (\frac{1mol}{17.03g}) ( \frac{4molNO}{4molNH_{3}}) = 0.147 mol NO
2.85g O_{2} (\frac{1mol}{32g}) ( \frac{4molNO}{5molO_{2}}) = 0.071 mol NO[/tex]
Step 5:
0.147 mol NO > 0.071 mol NO
Therefore, O₂ ang limiting reagent and NH₃ ang excess.
