Question 1
Why must a theoretical yield use the limiting reactant?
Need a hint for Question 1?
Use the relationships above and check particle ratios and units.
Check Question 1
That reactant runs out first and sets the maximum amount of product.
STUDENT SPACE · TOPIC 2
Reacting moles and limiting reactants
Use mole ratios to calculate product mass, limiting reactants and yield.
The essentials
- 2Mg + O₂ → 2MgO gives a mole ratio of 2:1:2.
- Convert mass to moles before applying the coefficient ratio.
- 4.8 g Mg in excess oxygen: n(Mg) = 4.8/24 = 0.20 mol.
- n(MgO) = 0.20 mol; mass = 0.20 × 40 = 8.0 g.
Key vocabulary
| English | Chinese |
|---|---|
| mole ratio | 物质的量之比 |
| limiting reactant | 限制反应物 |
| percentage yield | 产率 |
Quick practice
Try each question before opening its hint or answer. Use paper for calculations and diagrams.
Question 2
2Mg + O₂ → 2MgO. Find the theoretical MgO mass from 2.4 g Mg and 3.2 g O₂. M: Mg = 24, O₂ = 32, MgO = 40 g/mol.
Need a hint for Question 2?
Calculate both reactant amounts, then compare with the 2:1 ratio.
Check Question 2
n(Mg) = 0.10; n(O₂) = 0.10 mol. Mg limits: 0.10 mol MgO = 4.0 g.
Question 3
The theoretical yield of MgO is 4.0 g. Calculate percentage yield if 3.0 g of dry MgO is collected.
Need a hint for Question 3?
Divide actual yield by theoretical yield, then multiply by 100.
Check Question 3
Yield = 3.0/4.0 × 100 = 75%.
Use Lesson 2 in your worksheet for written practice. Your teacher will discuss those answers in class.