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Objectives
Objectives
Calculate stoichiometric reacting masses involving limiting reactants
A reminder to please only use your IGCSE periodic table for stoichiometry.
Mass and Molar Mass calculations
Mass and Molar Mass calculations
You must memorize this triangle. It is quicker with the symbols if you can.
Mass, m (g)
Molar Mass, M (g/mol)
Number of moles, n (mol)
Limiting reactants
Limiting reactants
Another excellent video. Useful if you missed the lesson.
Limiting reactants
Limiting reactants
Consider this balanced combustion reaction:
magnesium + oxygen → magnesium oxide
2Mg + O2 → 2MgO
So 2 moles of magnesium metal would combust completely in 1 mole of oxygen gas.
Question:
Question:
What would happen if we combusted 2 moles of magnesium in 5 moles of oxygen?
Answer
Answer
All of the 2 moles of magnesium will have completely combusted in the first 1 mole of oxygen gas. Therefore the remaining 4 moles of oxygen gas will remain unreacted. We can say that:
The magnesium is the limiting reactant (limiting reagent)
The oxygen is in excess
Practice question with moles only
Practice question with moles only
Zinc reacts with hydrochloric acid (HCl) to produce zinc chloride and hydrogen gas. What is the limiting reactant if 0.5 mol of Zn and 0.4 mol of HCl are used?
Write word and symbol equation
Balance equation. Write the number of moles of both reactants.
Compare the available moles of each reactant to the moles required for a complete reaction using the molar ratio by scaling the equation.
(try starting with 0.5 mol Zn OR try starting with 0.4 mol HCl)
Answer
Answer
Zinc reacts with hydrochloric acid (HCl) to produce zinc chloride and hydrogen gas. What is the limiting reactant if 0.5 mol of Zn and 0.4 mol of HCl are used?
Write word and symbol equation
zinc + hydrochloric acid → zinc chloride + hydrogen
Zn + HCl → ZnCl2 + H2
Balance equation. Write the number of moles of both reactants.
Zn + 2HCl → ZnCl2 + H2
n=0.5mol n=0.4mol
Compare the available moles of each reactant to the moles required for a complete reaction using the molar ratio by scaling the equation.
If all of the 0.5 mol of Zn were used in the reaction, how many moles of HCl would be needed?
Zn + 2HCl → ZnCl2 + H2
↓x0.5 ↓x0.5
0.5Zn 1HCl
We would need 1 mol of HCl, but we only have 0.4 mol of HCl.
If all of the 0.4 mol of HCl were used in the reaction, how many moles of Zn would be needed?
Zn + 2HCl → ZnCl2 + H2
↓x0.2 ↓x0.2
0.2Zn 0.4HCl
We would need 0.2 mol of Zn. We have 0.5 mol of Zn, which is more than needed.
ANSWER: The HCl is the limiting reactant.
Practice question with masses
Practice question with masses
Let’s use the previous question but change it slightly. Zinc reacts with hydrochloric acid (HCl) to produce zinc chloride (ZnCl2) and hydrogen gas. What is the limiting reactant if 10g of Zn and 8.0g of HCl are used?
Write word and symbol equation
Balance equation. Write the number of moles of both reactants.
Compare the available moles of each reactant to the moles required for a complete reaction using the molar ratio by scaling the equation.
(try starting with 10g Zn OR try starting with 8.0g HCl)
Answer
Answer
Write word and symbol equation
zinc + hydrochloric acid → zinc chloride + hydrogen
Zn + HCl → ZnCl2 + H2
Balance equation. Write the number of moles of both reactants.
Zn + 2HCl → ZnCl2 + H2
m=10g m=8.0
n=0.154mol n=0.219mol
Compare the available moles of each reactant to the moles required for a complete reaction using the molar ratio by scaling the equation.
If all of the 0.154 mol of Zn were used in the reaction, how many moles of HCl would be needed?
Zn + 2HCl → ZnCl2 + H2
↓x0.154 ↓x0.154
0.154Zn 0.308HCl
We would need 0.308 mol of HCl, but we only have 0.219 mol of HCl.
If all of the 0.219 mol of HCl were used in the reaction, how many moles of Zn would be needed?
Zn + 2HCl → ZnCl2 + H2
↓x0.219 ↓x0.219
0.110Zn 0.219HCl
We would need 0.110 mol of Zn. We have 0.154 mol of Zn, so Zn is in excess.
ANSWER: The HCl is the limiting reactant.
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