Excess and Limiting Reagents: Concepts and Calculations

Limiting Reagent: The reactant that is completely consumed in a reaction, thus limiting the amount of product formed.

It is the reactant that determines the maximum amount of product that can be formed.

Generally when the limiting reagent is completely used used up, the reaction stops, that is, no further reaction proceeds.

Excess Reagent: The reactant that remains after the reaction is complete.

After the limiting reagent is used up, the excess reagent remains in the reaction mixture.

The presence of an excess reagent ensures that the limiting reagent is fully utilized, and the maximum possible amount of product is formed. However, it does not contribute to overall yield of the reaction.

Numerical Problems on Excess and Limiting Reagents

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  1. If 10g of magnesium (Mg) reacts with 20g of hydrochloric acid (HCl) according to the balanced equation Mg + 2HCl →MgCl2 + H2,
    (a)which reagent is the limiting reagent and
    (b)how much of the excess reagent remains after the reaction is complete?

    2. Solution
    (i) Determine the excess/limiting reagent
    Mg + 2HCl → MgCl2 + H2
    From the equation:
    1 mole of Mg requires 2 mole of HCl
    But 
    Mole = mass
       molar mass
    Mass of Mg = 1mol x 24g/mol = 24g
    Mass of HCl = 2mol x 36.5g/mol = 73g
    Therefore,
    From the equation:
    24g of Mg →73g of HCl
    10g of Mg →? g of HCl 

    Mass of HCl = 10g X 73g 
                24g  
	    = 30.42g
    Therefore, 10g of Mg would require 30.42g of HCl. Unfortunately there are only 20g, therefore, HCl is the limiting reagent.
    (b) To determine the amount of the excess reagent remained:
    Let us know how much of the amount of excess reagent reacted with the limiting reagent available.
    From the equation:
    24g of Mg → 73g of HCl
    ? g of Mg → 20g of HCl 
    Mass of Mg =   24g x 20g
                   73g
           = 6.58g (This is the amount of excess Mg reacted with HCl available)
    Therefore the amount of excess Mg remained = 10g - 6.58g = 3.42g
  2. (a) How many moles of nitrogen monoxide gas (NO) are produced when 2.5 moles of ammonia (NH3) react with 1.5 moles of oxygen gas (O2) according to the balanced equation 4NH3 + 5O2 → 4NO + 6H2O?
    (b)Which reagent in 2(a) above is the limiting reagent and how many moles of the excess reagent remain after the reaction is complete?

    3. Solution
    (i) Use stoichiometry relationship to determine the excess/limiting reagent:
    4NH3 + 5O2 → 4NO + 6H2O
    4 moles of NH3 → 5 moles of O2
    2.5 moles of NH3 → ? moles of O2
    Moles of O2 = 2.5mol x 5mol  
                  4mol
	   = 3. 125mol
    (a)The amount of product is determined by the limiting reagent, for this case is oxygen.
    It follows that, 
    5 moles of O2 → 4 moles of NO
1.5 moles of O2 → ? moles of NO

Moles of NO = 1.5mol x 4mol 
                 5 mol
            = 1.2 moles
1.2 moles of NO will be produced.
    (b)(i) Oxygen is the limiting agent.
    (ii) To calculate the amount of excess reagent remained:
     5 moles of O2 →  4 moles of NH3 
 1.5 moles of O2 → ? moles of NH3
  
  Moles of NH3 = 1.5mol x 4 mol
                     5 mol
               = 1.2 mol
These 1.2 moles of NH3 reacted with oxygen. 
    
→ 2.5 moles - 1.2 moles = 1.3 moles remained
  3. A chemist wants to react 25 g of sulfur (S) with 50 g of oxygen gas (O2) to produce sulfur dioxide (SO2) according to the balanced equation S + O2 → SO2.
    (i) Which reagent is the limiting reagent and
    (ii) how much of the excess reagent remains after the reaction is complete?

    4. Solution
    S + O2 → SO2.
    mass of S=32g, mass of O2 = 32g
    (i) to determine the limiting reagent:
    32g of S → 32g of O2
25g of S → ? g of O2
    
   mass of O2 =  25g x 32g
                  32g
	      = 25g
    Therefore oxygen is an excess reagent.
    ii. The amount of excess reagent remained:
    32g of S → 32g O2
 25g of S → ?g O2
Mass of S = 25g x 32g
	      32g	  
	 = 25g of oxygen
        50g - 25g = 25g
The amount of excess reagent remained is 25g.
  4. Sodium reacts with chlorine gas according to the equation 2Na + Cl2 → 2NaCl. If 10g of sodium (Na) were allowed to react with 20 g of chlorine gas (Cl2)
    (a) How many grams of sodium chloride (NaCl) are produced?
    (b) Which reagent is the limiting reagent
    (c)how many grams of the excess reagent remain after the reaction is complete?
    5. ( Relative atomic masses, Cl = 35.5, Na=23 )
    Solution
    2Na + Cl2 → 2NaCl
    (i) Determine excess/limiting reagents:
    From the equation: Mass of Na = 2mol x 23g/mol = 46g, 
                  Mass of chlorine = 1mol x 71g/mol = 71g
Therefore, 46g of Na → 71g of chlorine
           10g of Na → ?g of chlorine
Mass of chlorine =  10g x 71g
                        46g
                 = 15.43g
Therefore according to the data given, chlorine gas is in excess.
(a) The amount of product is determined by the limiting reagent:
    
    Therefore:
	Mass of NaCl = 2mol x (23 + 35.5)g/mol = 117g (... from the equation)
	 46g of Na → 117g of sodium chloride
	 10g of Na → ? g of sodium chloride
	 
	    Mass of sodium chloride          =  10g x 117g
					           46g
					     = 25.43g of sodium chloride
	Therefore, 25.43g of sodium chloride are produced.	
(b) Sodium is the limiting reagent	
(c) Excess reagent remained is:
    
     = 20g - 15.43g (of chlorine)
     = 4.57g
  5. A student wants to react 25 mL of 0.2 M hydrochloric acid (HCl) with 50 mL of 0.1 M sodium hydroxide (NaOH) to produce water (H2O) and sodium chloride (NaCl) according to the balanced equation HCl + NaOH -> NaCl + H2O. Which reagent is the limiting reagent and how many moles of the excess reagent remain after the reaction is complete?

    6. Solution
    (i) Determine the number of moles of reagents:
    
Volumes: → HCl 25mL = 0.025L, NaOH 50mL = 0.05L
Molarity = number of moles
              volume in L
Therefore, number of moles = molarity x volume(dm3)
  For hydrochloric acid: 0.2mol/dm3 x  0.025dm3  
                        = 0.005mol
  For sodium hydroxide:  0.1mol/dm3 x 0.05dm3	
                      = 0.005mol  
Their mole ratio equals to 1:1
 (a) Limiting reagent: neither (both reacted completely)
 (b) The excess reagent remained: None
  6. How many grams of carbon dioxide (CO2) are produced when 50.0 g of propane (C3H8) react with 100.0 g of oxygen gas (O2) according to the balanced equation C3H8 + 5O2→ 3CO2 + 4H2O? Which reagent is the limiting reagent and how many grams of the excess reagent remain after the reaction is complete?
    7. (Use the following relative atomic masses: C = 12, H =1 )
    Solution
    C3H8 + 5O2 → 3CO2 + 4H2O
Molar mass of propane = (12 x 3) + (1 x 8) = 44g/mol 
Molar mass of oxygen = 2 x 16 = 32g/mol 

From the equation:
Mass of propane = 1mol x 44g/mol = 44g
Mass of oxygen = 5mol x 32g/mol = 160g

Stoichiometry:
44g of propane requires 160g of oxygen
50g of propane requires ?  g of oxygen

Mass of oxygen gas = 50g x 160g
                       44g			 
                   = 181.82g
This is the mass of oxygen required to react with 50g of propane, unfortunately there are only 100.0g.
 Oxygen is the limiting reagent.   

 To calculate the mass of carbon dioxide :
The molar mass of carbon dioxide = 44g/mol
Mass of carbon dioxide from the equation = 3mol x 44g/mol
                                        = 132g
									  
Stoichiometry:
160g of oxygen were required in order 132g of carbon dioxide to be produced.
100g of oxygen will produce ? g of carbon dioxide 

           =  100g x 132g
	         160g
	   = 82.5g
Reagent remained:   
160g of oxygen → 44g of propane
100g of oxygen → ? g of propane	
	 
Mass of propane = 100g x 44g 
                      160g	
                = 27.5g		

  50.0 - 27.5 = 22.5g	
Therefore, the mass of reagent remained after the reaction were 22.5g.
  7. Methane (CH₄) reacts with chlorine gas (Cl₂) to produce carbon tetrachloride (CCl₄) and hydrogen chloride (HCl). If the chemical equation for the reaction is:
    CH₄ + 4Cl₂ → CCl₄ + 4HCl
    The chemical reactants present are 10 grams of CH₄ and 100 grams of Cl₂, which is the limiting reagent?
  8. Ammonia (NH₃) reacts with oxygen (O₂) to produce nitrogen (N₂) and water (H₂O). The summary for this reaction is:
    4NH₃ + 3O₂ → 2N₂ + 6H₂O
    If you have 50 grams of NH₃ and 50 grams of O₂, what will be the limiting reagent in this reaction?
  9. In a reaction, 10 grams of hydrogen gas (H2) and 15 grams of oxygen gas (O2) are combined. If the balanced equation requires 2 moles of H2 and 1 mole of O2, which reactant is the limiting reagent?
    Answer: Oxygen gas (O2) is the limiting reagent.
  10. If 20 grams of methane (CH4) and 30 grams of carbon dioxide (CO2) are combined, and the balanced equation requires 1 mole of CH4 and 1 mole of CO2, how many grams of the excess reagent will be left over after the reaction is complete?
    Answer: 16 grams of carbon dioxide (CO2) will be left over.
  11. You are a chemical engineer working in a manufacturing plant that produces ammonia (NH₃) using the Haber process. The plant needs to produce 850 grams of ammonia for an industrial order. You have a reactor that is currently loaded with 500 grams of hydrogen gas (H₂) and 1000 grams of nitrogen gas (N₂). Your task is
    1. determine the quantities of reactants required
    2. identify the limiting reagent to optimize the production process.
    3. determine if the plant can meet the production requirement.
    4. calculate the amount of excess reagent left over after the reaction.
    The chemical equation for the Haber process is:
    N2 + 3H2 → 2NH3

    12. Solution
    1. To determine the quantities of reactants required:
      From the equation:
Masses of nitrogen: 1mol x 28g/mol = 28g
Mass of hydrogen : 3mol x (1 x 2)g/mol = 6g
Mass of ammonia : 2mol x (14 +  1x3) = 34g
Therefore, 
Nitrogen gas needed:
28g of nitrogen gas → 34g of ammonia
? g of nitrogen gas → 850g of ammonia

  Mass of nitrogen gas = 28g x 850g
                           34g
		       = 700g
			   
Hydrogen gas needed:
6g of hydrogen gas → 34g of ammonia 
? g of hydrogen gas → 850g of ammonia

Mass of hydrogen gas = 6g x 850g 
                          34g
		     = 150g
    2. Identifying the limiting reagent: 
      From the equation 
28g of nitrogen gas → 6g of hydrogen gas
1000g of nitrogen gas → ?g of hydrogen gas
Mass of hydrogen gas = 1000g x 6g
                        28g
				= 214.3g 
Therefore, nitrogen is the limiting reagent.
    3. to determine if the plant can meet the production requirement. 
      28g of nitrogen gas → 34g of ammonia 
1000g of nitrogen gas → ?g of ammonia

mass of ammonia = 1214.3g
Since the grams of ammonia that can be 
      produced by a limiting agent is higher than 
      the order, therefore the plant can meet 
      the production requirement.
    4. To determine the amount of excess reagent left over after the reaction. 
      The remaining excess reagent (H2:
    500g - 214.3g = 285.7g
  12. Glucose (C₆H₁₂O₆) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) according to equation
    C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.
    If 180 grams of glucose and 192 grams of oxygen were mixed together and allowed to react, which is the limiting reagent? How much carbon dioxide is produced?
  13. In a reaction where 16 grams of Oxygen (O2) react with 12 grams of Hydrogen (H2) to form water (H2O), determine which substance is the limiting reagent and calculate the maximum amount of water that can be produced. (Hint: Use the relative atomic masses of O=16, H=1)
  14. A chemist mixes 140 grams of sodium chloride (NaCl) with 100 grams of silver nitrate (AgNO3) in a double displacement reaction to produce sodium nitrate (NaNO3) and silver chloride (AgCl). Identify the limiting reagent and calculate how much silver chloride is produced.
  15. A chemical reaction requires 9 grams of magnesium (Mg) and 32 grams of sulphur (S) to produce magnesium sulphide (MgS). Determine the excess reagent and calculate how much magnesium sulphide is formed.
  16. In an industrial process, 300 grams of carbon (C) and 400 grams of oxygen (O2) are combined to produce carbon dioxide (CO2). Which substance will be the limiting reagent, and what is the maximum amount of carbon dioxide that can be produced?
  17. A student performs a reaction involving 15 grams of hydrogen (H2) and 100 grams of nitrogen (N2) to produce ammonia (NH3). Determine the limiting reagent and calculate the amount of ammonia produced.
  18. For a reaction producing sodium hydroxide (NaOH) from sodium (Na) and water (H2O), 23 grams of sodium (Na) and 18 grams of water (H2O) are used. What is the limiting reagent, and calculate the amount of sodium hydroxide produced.
  19. In the anodizing process 15.0 grams of aluminium (Al) and 20.0 grams of oxygen (O₂) were allowed to react.The chemical equation for the reaction is:
    4Al + 3O2 → 2Al2O3.
    1. Identify the limiting reagent.
    2. Calculate the amount of aluminum oxide (Al₂O₃) produced.
    3. Determine the amount of excess reagent remaining.
  20. Three moles of nitrogen gas combine with five moles of hydrogen gas to form ammonia gas by Haber process.
    1. Which reactant is present in smaller amount?
    2. Calculate the grams of the reactant left in the container.
    3. How many moles of NH3 are produced?
    4. How many litres of NH3 are produced at S.T.P? (NECTA, 2018)

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Reflection
  1. What is a limiting reagent in a chemical reaction?
    The limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product that can be formed.
  2. How do you determine the limiting reagent in a chemical reaction?
    To determine the limiting reagent, you need to compare the moles or quantities of each reactant to the stoichiometric ratio in the balanced equation.
  3. What is the purpose of excess reagent in a chemical reaction?
    The purpose of excess reagent is to ensure that the limiting reagent is completely consumed, allowing for the maximum possible amount of product to be formed.
  4. How can you identify the excess reagent in a chemical reaction?
    The excess reagent is the reactant that remains after the limiting reagent is completely consumed in a chemical reaction.
  5. What happens to the excess reagent after a chemical reaction is complete?
    After a chemical reaction is complete, the excess reagent remains in the reaction mixture and can be separated from the desired product.
  6. Why is it important to identify the limiting reagent in a chemical reaction?
    It is important to identify the limiting reagent because it determines the maximum amount of product that can be formed in a chemical reaction.
  7. How does the amount of limiting reagent affect the amount of product formed in a chemical reaction?
    The reaction will stop once the limiting reagent is completely consumed, regardless of the excess reagent present.

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