Question: 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°C 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
Calculate moles of CO(g) available to react n(CO(g)) = V(CO(g)) ÷ Vm At STP. 1 mole of gas has a volume of 22.71 L
Vm = 22.71 L mol-1
V(CO(g)) = 45.42 Lmoles of CO = 45.42 ÷ 22.71 = 2 mol
Calculate moles of O2(g) available to react n(O2) = V(O2(g) ÷ Vm At STP 1 mole of gas has a volume of 22.71 L
Vm = 22.71 L mol-1
V(O2(g)) = 11.36 Lmoles O2 = 11.36 ÷ 22.71 = 0.5 mol
- Use the balanced chemical equation to determine the mole ratio (stoichiometric 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 × 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 × 0.5 = 1 mole of CO.
There are 2 moles of CO available which is more than the required 1 mole.
- (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.
(ii) 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)
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which of the following is not true about limiting and excess reagents? a. the amount of product obtained is determined by the limiting reagent. b. a balanced equation is necessary to determine which reactant is limiting reagent. c. some of the excess reagent is left over after the reaction is complete. d. the reactant that was the smallest given mass is the limiting reagent.
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| Limiting reagent•Whenever two or more reactants are given in a chemical reaction, the limiting reactant must be identified before you can calculate the maximum amount of product that can be produced.•A limiting reactant is completely used up in the chemical reaction.•The other reactants are known as excess reactants and are not completely used up. Sample Problem: •6.00g of magnesium reacts with 2.00g of oxygen to form magnesium oxide.•Determine the limiting reactant, the excess reactant (how many mole in excess) & mass of magnesium oxide formed.Solution: Step 1: Write a balanced equation and identify known and unknown. 2Mg (s) + O2 (g) → 2MgO(s) known known unknown 6.00 g 2.00 g Step 2: Calculate n (known). n(Mg) = n / M = 6.00 / 24.3 = 0.247 mol n(O2) = n / M = 2.00 / 32.0 = 0.0625 molCompare mole ratio from equation: Mg : O2 2 : 1 0.247 : 0.1235 (required)Compare mole of O2 required to actual.•Compare mole of O2 required to actual. Required : Actual 0.1235 : 0.0625•Because the actual mole is less than the required mole, there is not enough O2.Therefore O2 is the limiting reactant.•Therefore Mg is the excess reactant.•To calculate the mole of Mg in excess use equation to compare mole ratios. Mg : O2 2 : 1 0.125 : 0.0625 •0.125 mole of Mg reacts with all of the O2•n (Mg) in excess = n (Mg) initial – n (Mg) reacted = 0.247 – 0.125 = 0.122 molStep 3: From equation find ratio of n(unknown) to n(known). n(MgO) = 2/1 x n(O2) = 2/1 x 0.0625 = 0.125 molStep 4: Answer Question Find mass of magnesium oxide. m (MgO) = n x M = 0.125 x (24.3 + 16.0) = 0.125 x 40.3 = 5.04 g | More on limiting reagents Kahoot quiz- Excess and limiting reagents Excess and limiting reagents |
Key Terms
Skills to Develop
- Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction and explain why.
- Calculate theoretical yields of products formed in reactions that involve limiting reagents.
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Let us consider the reaction between solid sodium and chlorine gas. The reaction can be represented by the equation:
\[\mathrm{2 Na_{\large{(s)}} + Cl_{2\large{(g)}} \rightarrow 2 NaCl_{\large{(s)}}}\]
It represents a reaction of a metal and a diatomic gas chlorine. This balanced reaction equation indicates that two \(\ce{Na}\) atoms would react with two \(\ce{Cl}\) atoms or one \(\ce{Cl2}\) molecule. Thus, if you have 6 \(\ce{Na}\) atoms, 3 \(\ce{Cl2}\) molecules will be required. If there is an excess number of \(\ce{Cl2}\) molecules, they will remain unreacted. We can also state that 6 moles of sodium will require 3 moles of \(\ce{Cl2}\) gas. If there are more than 3 moles of \(\ce{Cl2}\) gas, some will remain as an excess reagent, and the sodium is a limiting reagent. It limits the amount of the product that can be formed.
Chemical reactions with stoichiometric amounts of reactants have no limiting or excess reagents.
Example \(\PageIndex{1}\)
Calculate the number of moles of \(\ce{CO2}\) formed in the combustion of ethane \(\ce{C2H6}\) in a process when 35.0 mol of \(\ce{O2}\) is consumed.
HINT
The reaction is
\[\ce{2 C2H6 + 7 O2 \rightarrow 4 CO2 + 6 H2O}\]
\[\mathrm{35.0\: mol\: O_2 \times\dfrac{4\: mol\: CO_2}{7\: mol\: O_2} = 20.0\: mol\: CO_2}\]
DISCUSSION
A balanced equation for the reaction is a basic requirement for identifying the limiting reagent even if amounts of reactants are known.
Example \(\PageIndex{2}\)
Two moles of \(\ce{Mg}\) and five moles of \(\ce{O2}\) are placed in a reaction vessel, and then the \(\ce{Mg}\) is ignited according to the reaction
\(\mathrm{Mg + O_2 \rightarrow MgO}\).
Identify the limiting reagent in this experiment.
HINT
Before a limiting reagent is identified, the reaction must be balanced. The balanced reaction is
\[\mathrm{2 Mg + O_2 \rightarrow 2 MgO}\]
Thus, two moles of \(\ce{Mg}\) require only ONE mole of \(\ce{O2}\). Four moles of oxygen will remain unreacted. Therefore, oxygen is the excess reagent, and \(\ce{Mg}\) is the limiting reagent.
DISCUSSION
Answer these questions: How many moles of \(\ce{MgO}\) is formed?
What is the weight of \(\ce{MgO}\) formed?
- At room temperature (25 °C) what is the state of sodium: solid, gas or liquid?
Hint: Sodium metal
- Equal weights of \(\ce{H2}\) and \(\ce{O2}\) are placed in a balloon and then ignited. Assuming reaction goes to completion, which gas is the excess reagent?
Hint: hydrogen
Hint: Iron oxide
Discussion -
A stoichiometric mixture has a mass ratio of 54:160 (nearly 1:3) for \(\ce{Al:Fe2O3}\).