• Next question: – When chemical reaction is complete, how much excess reagent is left??? Excess Reactant
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• Next question: – When chemical reaction is complete, how much
excess reagent is left???
Excess Reactant
The Sandwich… again…
You have:10 slices of bread3 ham
After using up all LR (ham), we know there are 4 slices of bread left.
Is there any “formula” we can use to find that?
How much of the excess reactant is left?
2 breads + 1 ham à 1 sandwich
• Amount of bread left =Starting amount of bread – Amount of bread used
• How can we use stoichiometry to figure out the amount of bread used?
10 (Given) 6 (How?)
3 ham = 6 breads×2 breads
1 ham
How much of the excess reactant is left?
CH4+Cl2→ CH3Cl + HCl• Amount of excess left =Starting mass of CH4 – mass of CH4 used
• Based on the mass of LR given, find the mass of excess reactant used.
• Cl2 g (10.0g) à Cl2 mol à CH4 mol à CH4 g
15.0g (Given) ?
How much of the excess reactant is left?
• The mass of ER used is proportional to the mass of LR used. So to find the mass of ER used, do another stoichiometry calculation, starting with the mass of LR given.
• 10.0g Cl2 à Cl2 mol à CH4 mol à CH4 g
15g 10g 7.11g
CH4+Cl2→ CH3Cl + HClExcess Limiting
Limiting reactant
Limiting reactant
Excess reactant actually
used
Limiting reactant
Excess reactant actually
used
Excess reactant actually used
16.0g CH4
1 mol CH4
1 mol Cl2
71.0 g Cl2
1 mol CH4
1 mol Cl2
CH4+Cl2→ CH3Cl + HClMass of CH4used =
10.0g Cl2
=2.253521127…g CH4 used
× × ×
MM(CH4)=16.0g/molMM(Cl2)= 71.0g/molAmount of excess CH4 left =
The initial mass of CH4 – The amount of CH4 used=
15.0 g(1DP) CH4 – 2.253521127…g(2DP) CH4 used = 12.74648…g CH4
= 12.7g CH4
How much of the excess reactant is left?
• Quick practice: – Calculate how much H2O can be made using
10.0g of H2 and 50.0g of O2
10.0g 50.0g 56.3g2H2 + O2 à 2H2O
Excess Limiting
– Recall: H2 is the ER and O2 is the LR– How many gram of H2 is left?
Remember this?
18.0g H2O
1 mol H2O
1 mol H2
2.0 g H2
2 mol H2O
2 mol H2
2H2+O2→ 2H2OMass of H2O produced (From H2)=
Quick Practice
10.0g H2
= 90.0g H2O
MM(H2)=2.0g/molMM(O2)=32.0g/molMM(H2O)= 18.0g/mol
× × ×
18.0g H2O
1 mol H2O
1 mol O2
32.0 g O2
2 mol H2O
1 mol O2
Mass of H2O produced (From O2)=
50.0g O2
= 56.3g H2O
× × ×
O2 is the LR and H2 is the ER
2.0g H2
1 mol H2
1 mol O2
32.0 g O2
2 mol H2
1 mol O2
2H2+O2→ 2H2OMass of H2 used=g O2 à mol O2 à mol H2 à g H2
Quick Practice
50.0g O2
= 6.25 g H2 used
MM(H2)=2.0g/molMM(O2)=32.0g/molMM(H2O)= 18.0g/mol
× × ×
Mass of H2 used = Starting mass – Used mass=10.0g H2 (1DP) – 6.25g H2 (1DP)
=3.8g H2 left (1DP)
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