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ChemistryEdu Logo 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

\[Moles\ of\ S = {10 \over 32}\]
\[Moles\ of\ O_2 = {10 \over 32}\]

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.

\[1\ mole\ of\ S\ forms\ 1\ mole\ of\ SO_2.\]
\[{10 \over 32} moles\ of\ S\ forms\ {10 \over 32} moles\ of\ SO_2.\]
\[Mass\ of\ SO_2 = {10 \over 32} \times 64 = 20\ g\]

For the given chemical reaction, find the mass of CaCl2 formed:
Ca(10 g) + Cl2(10 g) → CaCl2

\[Moles\ of\ Ca = {10 \over 40}\]
\[Moles\ of\ Cl_2 = {10 \over 71}\]

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

\[1\ mole\ of\ Cl_2\ forms\ 1\ mole\ of\ CaCl_2.\]
\[{10 \over 71} moles\ of\ Cl_2\ forms\ {10 \over 71} moles\ of\ CaCl_2.\]
\[Mass\ of\ CaCl_2 = {10 \over 71} \times 111 = {1110 \over 71}\ g\]

For the given chemical reaction, find the mass of PCl5 formed:
P4(10 g) + 10Cl2(10 g) → 4PCl5

\[Moles\ of\ P_4 = {10 \over 124}\]
\[Moles\ of\ Cl_2 = {10 \over 71}\]

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

\[10\ moles\ of\ Cl_2\ form\ 4\ moles\ of\ PCl_5.\]
\[1\ mole\ of\ Cl_2\ forms\ {4\over 10} moles\ of\ PCl_5.\]
\[{10 \over 71} moles\ of\ Cl_2\ forms\ {4 \over 10} \times {10 \over 71} moles\ of\ PCl_5.\]
\[Mass\ of\ PCl_5 = {4 \over 10} \times {10 \over 71} \times 208.5 = {834 \over 71}\ g\]