Mole Concept | Limiting Reagent#

Limiting Reagent#

Limiting reagent is the reactant which gets completely consumed in a chemical reaction. It limits the amount of products formed.
Limiting reagent does not refer to the reagent present in lesser quantity. Even the reagent present in larger quantity may act like limiting reagent.

How to identify Limiting Reagent?#

Divide the number of moles of reactants by their respective stoichiometric coefficient. The lesser value will indicate that the reactant is limiting reagent.
After identifying limiting reagent, make all calculations with respect to limiting reagent only.

Questions#

For the given chemical reaction, find the mass of SO₂ formed:
S_{(10 g)} + O₂_{(10 g)} → SO₂

M o l e s o f S = \frac{10}{32}

M o l e s o f O_{2} = \frac{10}{32}

Dividing these moles by corresponding stoichiometric coefficient, we find that the values are equal for both S and O₂. Hence, both are limiting reagents. Both S and O₂ will be completely consumed to form SO₂.

1 m o l e o f S f o r m s 1 m o l e o f S O_{2} .

\frac{10}{32} m o l e s o f S f o r m s \frac{10}{32} m o l e s o f S O_{2} .

M a s s o f S O_{2} = \frac{10}{32} \times 64 = 20 g

For the given chemical reaction, find the mass of CaCl₂ formed:
Ca_{(10 g)} + Cl₂_{(10 g)} → CaCl₂

M o l e s o f C a = \frac{10}{40}

M o l e s o f C l_{2} = \frac{10}{71}

Dividing these moles by corresponding stoichiometric coefficient, we find that the value for both Cl₂ is less. Hence, Cl₂ is the limiting reagent.

1 m o l e o f C l_{2} f o r m s 1 m o l e o f C a C l_{2} .

\frac{10}{71} m o l e s o f C l_{2} f o r m s \frac{10}{71} m o l e s o f C a C l_{2} .

M a s s o f C a C l_{2} = \frac{10}{71} \times 111 = \frac{1110}{71} g

For the given chemical reaction, find the mass of PCl₅ formed:
P₄_{(10 g)} + 10Cl₂_{(10 g)} → 4PCl₅

M o l e s o f P_{4} = \frac{10}{124}

M o l e s o f C l_{2} = \frac{10}{71}

Dividing these moles by corresponding stoichiometric coefficient, we find that the value for both Cl₂ is less ( $\frac{\frac{10}{71}}{10} < \frac{10}{124}$ ). Hence, Cl₂ is the limiting reagent.

10 m o l e s o f C l_{2} f o r m 4 m o l e s o f P C l_{5} .

1 m o l e o f C l_{2} f o r m s \frac{4}{10} m o l e s o f P C l_{5} .

\frac{10}{71} m o l e s o f C l_{2} f o r m s \frac{4}{10} \times \frac{10}{71} m o l e s o f P C l_{5} .

M a s s o f P C l_{5} = \frac{4}{10} \times \frac{10}{71} \times 208.5 = \frac{834}{71} g