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Limiting Reagents & Reactants in Excess |
Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl
Zn + 2HCl -----> ZnCl2 + H2
| moles of Zn = 0.5 | moles of HCl = 0.4 |
| Zn : HCl | Or | HCl : Zn |
| 1 : 2 |   | 1 : ½ |
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 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.
Find the limiting reagent and the reactant in excess when 1.5g of CaCO3 react completely with 0.73g of HCl
CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O
| moles of CaCO3 = mass ÷ MM
mass = 1.5g MM = 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 ÷ MM
mass = 0.73g MM = 1.008 + 35.45 = 36.458g/mol moles HCl = 0.73 ÷ 36.458 = 0.02 mol |
| CaCO3 : HCl | Or | HCl : CaCO3 |
| 1 : 2 |   | 1 : ½ |
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 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.
Find the limiting reagent and the reactant in excess when 100mL of 0.2 NaOH react completely with 50mL of 0.5M H2SO4
2NaOH + H2SO4 -----> Na2SO4 + 2H2O
| moles of NaOH = M x V
M = 0.2M V = 100mL = 100 x 10-3L moles of NaOH = 0.2 x 100 x 10-3 = 0.02 mol |
moles of H2SO4 = M x V
M = 0.5M V = 50mL = 50 x 10-3L moles H2SO4 = 0.5 x 50 x 10-3 = 0.025mol |
| NaOH : H2SO4 | Or | H2SO4 : NaOH |
| 1 : ½ |   | 1 : 2 |
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 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.
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)
2CO(g) + O2(g) -----> 2CO2(g)
| moles of CO = V ÷ 22.41
At S.T.P. 1 mole of gas has a volume of 22.41L moles of CO = 44.82 ÷ 22.41 = 2mol |
moles of O2 = V ÷ 22.41
At S.T.P. 1 mole of gas has a volume of 22.41L moles O2 = 11.205 ÷ 22.41 = 0.5mol |
| CO : O2 | Or | O2 : CO |
| 1 : ½ |   | 1 : 2 |
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 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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