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Limiting Reagents & Reactants in Excess

Key Concepts

The limiting reagent is the reactant that is completely used up during the chemical reaction.
The reactant that is in excess is the reactant that is not completely used up during the chemical reaction, that is, there is some of this reactant left over.

Deciding which reactants are the limiting reagents and the reactants in excess

Write the balanced chemical equation for the chemical reaction
Calculate the available moles of each reactant in the chemical reaction
Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.

Examples

Moles of reactants given

Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

Write the balanced chemical equation for the chemical reaction
Zn + 2HCl → ZnCl₂ + H₂
Calculate the available moles of each reactant in the chemical reaction

moles of Zn = 0.5 moles of HCl = 0.4
Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

Zn : HCl Or HCl : Zn

1 : 2 1 : ½
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
If all of the 0.5 moles of Zn were to be used in the reaction it would require
2 x 0.5 = 1.0 moles of HCl for the reaction to go to completion.
There are only 0.4 moles of HCl available which is less than the required 1.0 moles.
If all of the 0.4 moles of HCl were to be used in the reaction it would require
½ x 0.4 = 0.2 moles Zn.
There are 0.5 moles of Zn available which is more than the required 0.2 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.
The limiting reagent is HCl,
all of the 0.4 moles of HCl will be used up when this reaction goes to completion.
The reactant in excess is Zn,
when the reaction has gone to completion there will be
0.5 - 0.2 = 0.3 moles of Zn left over.

Masses of reactants given

Find the limiting reagent and the reactant in excess when 1.5 g of CaCO₃ react completely with 0.73 g of HCl

Write the balanced chemical equation for the chemical reaction
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Calculate the available moles of each reactant in the chemical reaction

moles of CaCO₃ = mass ÷ molar mass
mass = 1.5 g
molar mass = 40.08 + 12.01 + (3 x 16.00)
= 100.09 g/mole

moles of CaCO₃ = 1.5 ÷ 100.09
= 0.015 mol

moles of HCl = mass ÷ molar mass
mass = 0.73g
molar mass = 1.008 + 35.45
= 36.458g/mol

moles HCl = 0.73 ÷ 36.458
= 0.02 mol

Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

CaCO₃ : HCl Or HCl : CaCO₃

1 : 2 1 : ½
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
If all of the 0.015 moles of CaCO₃ were to be used in the reaction it would require
2 x 0.015 = 0.03 moles of HCl for the reaction to go to completion.
There are only 0.02 moles of HCl available which is less than the required 0.03 moles.
If all of the 0.02 moles of HCl were to be used in the reaction it would require
½ x 0.02 = 0.01 moles of CaCO₃.
There are 0.015 moles of CaCO₃ available which is more than the required 0.01 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.
The limiting reagent is HCl,
all of the 0.02 moles of HCl will be used up when this reaction goes to completion.
The reactant in excess is CaCO₃,
when the reaction has gone to completion there will be
0.015 - 0.01 = 0.005 moles of CaCO₃ left over.

Concentration and volume of solutions given

Find the limiting reagent and the reactant in excess when 100 mL of 0.2 mol L^-1 NaOH aqueous solution react completely with 50 mL of 0.5 mol L^-1 H₂SO₄ aqueous solution.

Write the balanced chemical equation for the chemical reaction
2NaOH_(aq) + H₂SO₄_(aq) → Na₂SO₄_(aq) + 2H₂O_(l)

Calculate the available moles of each reactant in the chemical reaction:
n = c x V
n = moles of solute
c = concentration of solution in mol L^-1
V = volume of solution in L

n(NaOH) = c(NaOH) x V
c(NaOH) = 0.2 mol L^-1
V = 100 mL = 100 x 10^-3 L

n(NaOH) = 0.2 x 100 x 10^-3 = 0.02 mol

n(H₂SO₄) = c(H₂SO₄) x V
c(H₂SO₄) = 0.5 mol L^-1
V = 50 mL = 50 x 10^-3 L

n(H₂SO₄) = 0.5 x 50 x 10^-3 = 0.025 mol

Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

NaOH : H₂SO₄ Or H₂SO₄ : NaOH

1 : ½ 1 : 2
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
If all of the 0.02 moles of NaOH were to be used in the reaction it would require
½ x 0.02 = 0.01 moles of H₂SO₄ for the reaction to go to completion.
There are 0.025 moles of H₂SO₄ available which is more than the required 0.01 moles.
If all of the 0.025 moles of H₂SO₄ were to be used in the reaction it would require
2 x 0.025 = 0.05 moles of NaOH.
There are only 0.02 moles of NaOH available which is less than the required 0.05 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.
The limiting reagent is NaOH,
all of the 0.02 moles of NaOH will be used up when this reaction goes to completion.
The reactant in excess is H₂SO₄,
when the reaction has gone to completion there will be
0.025 - 0.01 = 0.015 moles of H₂SO₄ left over.

Gas Volumes given

Find the limiting reagent and the reactant in excess when 45.42 L of CO(g) react completely with 11.36 L of O₂(g) at STP (0^oC or 273.15 K and 100 kPa)

Write the balanced chemical equation for the chemical reaction
2CO(g) + O₂(g) → 2CO₂(g)

Calculate the available moles of each reactant in the chemical reaction

moles of CO = V ÷ V_m
At STP. 1 mole of gas
has a volume of 22.71L

moles of CO = 45.42 ÷ 22.71
= 2 mol

moles of O₂ = V ÷ V_m
At STP 1 mole of gas
has a volume of 22.71L

moles O₂ = 11.36 ÷ 22.71
= 0.5 mol

Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

CO : O₂ Or O₂ : CO

1 : ½ 1 : 2
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
If all of the 2 moles of CO were to be used in the reaction it would require
½ x 2 = 1 mole of O₂ for the reaction to go to completion.
There are 0.5 moles of O₂ available which is less than the required 1 mole.
If all of the 0.5 moles of O₂ were to be used in the reaction it would require
2 x 0.5 = 1 mole of CO.
There are 2 moles of CO available which is more than the required 1 mole.
The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.
The limiting reagent is O₂,
all of the 0.5 moles of O₂ will be used up when this reaction goes to completion.
The reactant in excess is CO,
when the reaction has gone to completion there will be
2 - 1 = 1 mole of CO left over.

Practice Questions

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