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 1.5g of CaCO3 react completely with 0.73g 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 ÷ MM
mass = 1.5g
MM = 40.08 + 12.01 + (3 x 16.00)
= 100.09 g/molemoles of CaCO3 = 1.5 ÷ 100.09
= 0.015 molmoles of HCl = mass ÷ MM
mass = 0.73g
MM = 1.008 + 35.45
= 36.458g/molmoles 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 100mL of 0.2 NaOH react completely with 50mL of 0.5M H2SO4
- Write the balanced chemical equation for the chemical reaction
2NaOH + H2SO4 -----> Na2SO4 + 2H2O
- Calculate the available moles of each reactant in the chemical reaction
moles of NaOH = M x V
M = 0.2M
V = 100mL
= 100 x 10-3Lmoles of NaOH = 0.2 x 100 x 10-3
= 0.02 molmoles of H2SO4 = M x V
M = 0.5M
V = 50mL
= 50 x 10-3Lmoles H2SO4 = 0.5 x 50 x 10-3
= 0.025mol - 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 44.82L of CO(g) react completely with 11.205L of O2(g) at S.T.P. (0oC or 273K and 1atm or 101.3kPa)
- 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 ÷ 22.41
At S.T.P. 1 mole of gas
has a volume of 22.41Lmoles of CO = 44.82 ÷ 22.41
= 2molmoles of O2 = V ÷ 22.41
At S.T.P. 1 mole of gas
has a volume of 22.41Lmoles O2 = 11.205 ÷ 22.41
= 0.5mol - 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.
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