Higher Tier: find which reactant limits the product
導入済み| English | 意味 |
|---|---|
| excess reactant/ekˈses rɪˈæktənt/ | 過剰反応物 |
| limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ | 限量反応物 |
What would explain this observation?
- Higher Tier: Having more grams of a reactant does not necessarily make it the excess reactant 过量反应物. The equation ratio and molar masses decide which supply runs out first.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- The limiting reactant 限量反应物 is completely used up under the complete-reaction model and sets the maximum product amount. An excess reactant remains because more was supplied than the balanced equation needs. Convert both reactant masses to amounts before comparing. For A + 2B → C, one mole A requires two moles B; compare n(A) with n(B)/2, not the raw mole amounts alone.
- limiting reactant: The reactant used up first under the stated reaction model, limiting possible product; excess reactant: A reactant supplied beyond the amount required to consume the limiting reactant.
For Mg + 2HCl → MgCl₂ + H₂, 0.20 mol Mg and 0.30 mol HCl are supplied. Which limits?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The equation needs 2 mol of HCl per mole of Mg. Available: 0.30 mol HCl can use only 0.15 mol of the 0.20 mol Mg, so HCl runs out first - it limits. Equal masses are irrelevant; the mole ratio decides.
Match each technical term to its precise meaning.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For Mg + 2HCl → MgCl₂ + H₂, 0.20 mol Mg would require 0.40 mol HCl. If only 0.30 mol HCl is supplied, acid limits and only 0.15 mol Mg can react. Maximum H₂ amount is 0.15 mol, and 0.05 mol Mg remains. This assumes reaction proceeds as written and does not describe incomplete conversion caused by slow rate or equilibrium.
- Write available amount, required coefficient and available amount divided by coefficient for each reactant. The smaller scaled supply determines the maximum reaction extent. Multiply that extent by the target product coefficient. To find excess remaining, subtract the amount consumed from the amount supplied. Avoid declaring a limiting reactant solely from a visual guess or whichever mass number is smaller.
Which two habits make the investigation or model in this case more defensible?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 4.8 g Mg is 4.8/24=0.20 mol. With 0.30 mol HCl, magnesium consumed=0.15 mol, mass=3.6 g. Remaining Mg=4.8−3.6=1.2 g. Hydrogen forms at 0.15 mol, mass=0.15×2=0.30 g using supplied molar mass H₂=2 $\dfrac{\text{g}}{\text{mol}}$.
For the same reaction, 0.40 mol Mg and 0.50 mol HCl are supplied. Calculate the maximum H₂ amount. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
For the same reaction, 0.40 mol Mg and 0.50 mol HCl are supplied. Calculate the maximum H₂ amount.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The result is 0.25 mol. Known: 0.40 mol Mg and 0.50 mol HCl; Mg + 2HCl -> MgCl2 + H2. HCl limits: 0.50 mol H2 produced per 2 mol HCl gives 0.50/2 = 0.25 mol of H2. Mg used = 0.25 mol, leaving 0.15 mol in excess.
Check the conclusion and its limits
- Excess means present beyond the required reacting amount, not an impurity. Adding more excess reactant alone does not raise the theoretical product after the limiting reactant is exhausted. Recovered yield can be lower than this maximum for other reasons; do not confuse limiting amount with percent recovery.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
The reactant with the smaller mass number must be limiting. This claim is false: Limiting is decided by the equation's mole ratio, not by mass or mass number. Divide each available amount by its coefficient; the smaller quotient limits, whatever the masses.
Higher Tier: find which reactant limits the product: For Mg + 2HCl → MgCl₂ + H₂, 0.20 mol Mg would require 0.40 mol HCl. If only 0.30 mol HCl is supplied, acid limits and only 0.15 mol Mg can react. Maximum H₂ amount is 0.15 mol, and 0.05 mol Mg remains. This assumes reaction proceeds as written and does not describe incomplete conversion caused by slow rate or equilibrium.
The reactant with the smaller mass number must be limiting.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Limiting is decided by the equation's mole ratio, not by mass or mass number. Divide each available amount by its coefficient; the smaller quotient limits, whatever the masses.
The reactant used up first under the stated reaction model, limiting possible product: write the technical term.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
limiting reactant means The reactant used up first under the stated reaction model, limiting possible product.