Detailed answer key for stoichiometry problems involving limiting reagents and excess reactants.
Chemistry worksheet showing stoichiometry limiting reagent problems with step-by-step answers and calculations.
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Step-by-step solution for: Limiting Reagent Worksheet Worksheet
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Step-by-step solution for: Limiting Reagent Worksheet Worksheet
Here are the step-by-step solutions for the three stoichiometry problems shown in your worksheet.
Reaction: $N_2 + 3 H_2 \rightarrow 2 NH_3$
Given: 28 g of $N_2$ and 25 g of $H_2$.
Part (a): How many grams of $NH_3$ can be produced?
To find the answer, we must determine which reactant is the "limiting reagent" (the one that runs out first). We do this by calculating how much product ($NH_3$) each reactant *could* make if it were used up completely.
1. Calculate potential $NH_3$ from Nitrogen ($N_2$):
* Molar mass of $N_2$ = $2 \times 14 = 28$ g/mol.
* Molar mass of $NH_3$ = $14 + (3 \times 1) = 17$ g/mol.
* Calculation:
$$28 \text{ g } N_2 \times \frac{1 \text{ mol } N_2}{28 \text{ g } N_2} \times \frac{2 \text{ mol } NH_3}{1 \text{ mol } N_2} \times \frac{17 \text{ g } NH_3}{1 \text{ mol } NH_3} = 34 \text{ g } NH_3$$
2. Calculate potential $NH_3$ from Hydrogen ($H_2$):
* Molar mass of $H_2$ = $2 \times 1 = 2$ g/mol.
* Calculation:
$$25 \text{ g } H_2 \times \frac{1 \text{ mol } H_2}{2 \text{ g } H_2} \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} \times \frac{17 \text{ g } NH_3}{1 \text{ mol } NH_3} \approx 141.67 \text{ g } NH_3$$
3. Compare:
* Nitrogen produces 34 g.
* Hydrogen produces ~142 g.
* Since 34 g is the smaller amount, Nitrogen is the limiting reagent, and the reaction stops there.
Answer for (a): 34 g of $NH_3$.
Part (b): How much of the excess reagent is left over?
The excess reagent is Hydrogen ($H_2$). To find what is left, we calculate how much was actually used to react with the Nitrogen, then subtract that from the starting amount.
1. Calculate $H_2$ used:
* We know we used all 28 g of $N_2$.
* Calculation:
$$28 \text{ g } N_2 \times \frac{1 \text{ mol } N_2}{28 \text{ g } N_2} \times \frac{3 \text{ mol } H_2}{1 \text{ mol } N_2} \times \frac{2 \text{ g } H_2}{1 \text{ mol } H_2} = 6 \text{ g } H_2 \text{ used}$$
2. Calculate remaining $H_2$:
* Start amount - Used amount = Remaining amount
* $25 \text{ g} - 6 \text{ g} = 19 \text{ g}$
Answer for (b): 19 g of $H_2$.
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Reaction: $Mg + 2 HCl \rightarrow MgCl_2 + H_2$
Given: 50 g of $Mg$ and 75 g of $HCl$.
Goal: Find the volume of $H_2$ gas produced at STP.
We need to find the limiting reagent again by seeing which reactant produces less hydrogen gas. At STP (Standard Temperature and Pressure), 1 mole of any gas equals 22.4 Liters.
1. Calculate potential $H_2$ from Magnesium ($Mg$):
* Molar mass of $Mg$ = 24.3 g/mol.
* Ratio: 1 mol $Mg$ makes 1 mol $H_2$.
* Calculation:
$$50 \text{ g } Mg \times \frac{1 \text{ mol } Mg}{24.3 \text{ g } Mg} \times \frac{1 \text{ mol } H_2}{1 \text{ mol } Mg} \times \frac{22.4 \text{ L } H_2}{1 \text{ mol } H_2} \approx 46.1 \text{ L } H_2$$
2. Calculate potential $H_2$ from Hydrochloric Acid ($HCl$):
* Molar mass of $HCl$ = $1 + 35.5 = 36.5$ g/mol.
* Ratio: 2 mol $HCl$ makes 1 mol $H_2$.
* Calculation:
$$75 \text{ g } HCl \times \frac{1 \text{ mol } HCl}{36.5 \text{ g } HCl} \times \frac{1 \text{ mol } H_2}{2 \text{ mol } HCl} \times \frac{22.4 \text{ L } H_2}{1 \text{ mol } H_2} \approx 23.0 \text{ L } H_2$$
3. Compare:
* Magnesium could make 46 L.
* HCl could make 23 L.
* The smaller amount is the correct answer because the HCl runs out first.
Answer: 23 L of $H_2$.
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Reaction: $Zn + S \rightarrow ZnS$
Given: 25 g of Zinc ($Zn$) and 30 g of Sulfur ($S$).
Goal: Identify the limiting reagent.
We compare how much product ($ZnS$) each reactant can make.
1. Calculate potential $ZnS$ from Zinc ($Zn$):
* Molar mass of $Zn$ = 65.4 g/mol.
* Molar mass of $ZnS$ = $65.4 + 32 = 97.4$ g/mol (using 97.48 as per sheet).
* Ratio: 1 mol $Zn$ makes 1 mol $ZnS$.
* Calculation:
$$25 \text{ g } Zn \times \frac{1 \text{ mol } Zn}{65.4 \text{ g } Zn} \times \frac{1 \text{ mol } ZnS}{1 \text{ mol } Zn} \times \frac{97.48 \text{ g } ZnS}{1 \text{ mol } ZnS} \approx 37.3 \text{ g } ZnS$$
2. Calculate potential $ZnS$ from Sulfur ($S$):
* Molar mass of $S$ = 32 g/mol.
* Ratio: 1 mol $S$ makes 1 mol $ZnS$.
* Calculation:
$$30 \text{ g } S \times \frac{1 \text{ mol } S}{32 \text{ g } S} \times \frac{1 \text{ mol } ZnS}{1 \text{ mol } S} \times \frac{97.48 \text{ g } ZnS}{1 \text{ mol } ZnS} \approx 91.4 \text{ g } ZnS$$
3. Compare:
* Zinc produces 37.3 g of product.
* Sulfur produces 91.4 g of product.
* Since Zinc produces the smaller amount, it limits the reaction.
Answer: Zn is the limiting reagent.
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Final Answer:
1a) 34 g NH₃
1b) 19 g H₂ remains
2) 23 L H₂
3) Zn is the limiting reagent
Problem 1: Nitrogen and Hydrogen Reaction
Reaction: $N_2 + 3 H_2 \rightarrow 2 NH_3$
Given: 28 g of $N_2$ and 25 g of $H_2$.
Part (a): How many grams of $NH_3$ can be produced?
To find the answer, we must determine which reactant is the "limiting reagent" (the one that runs out first). We do this by calculating how much product ($NH_3$) each reactant *could* make if it were used up completely.
1. Calculate potential $NH_3$ from Nitrogen ($N_2$):
* Molar mass of $N_2$ = $2 \times 14 = 28$ g/mol.
* Molar mass of $NH_3$ = $14 + (3 \times 1) = 17$ g/mol.
* Calculation:
$$28 \text{ g } N_2 \times \frac{1 \text{ mol } N_2}{28 \text{ g } N_2} \times \frac{2 \text{ mol } NH_3}{1 \text{ mol } N_2} \times \frac{17 \text{ g } NH_3}{1 \text{ mol } NH_3} = 34 \text{ g } NH_3$$
2. Calculate potential $NH_3$ from Hydrogen ($H_2$):
* Molar mass of $H_2$ = $2 \times 1 = 2$ g/mol.
* Calculation:
$$25 \text{ g } H_2 \times \frac{1 \text{ mol } H_2}{2 \text{ g } H_2} \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} \times \frac{17 \text{ g } NH_3}{1 \text{ mol } NH_3} \approx 141.67 \text{ g } NH_3$$
3. Compare:
* Nitrogen produces 34 g.
* Hydrogen produces ~142 g.
* Since 34 g is the smaller amount, Nitrogen is the limiting reagent, and the reaction stops there.
Answer for (a): 34 g of $NH_3$.
Part (b): How much of the excess reagent is left over?
The excess reagent is Hydrogen ($H_2$). To find what is left, we calculate how much was actually used to react with the Nitrogen, then subtract that from the starting amount.
1. Calculate $H_2$ used:
* We know we used all 28 g of $N_2$.
* Calculation:
$$28 \text{ g } N_2 \times \frac{1 \text{ mol } N_2}{28 \text{ g } N_2} \times \frac{3 \text{ mol } H_2}{1 \text{ mol } N_2} \times \frac{2 \text{ g } H_2}{1 \text{ mol } H_2} = 6 \text{ g } H_2 \text{ used}$$
2. Calculate remaining $H_2$:
* Start amount - Used amount = Remaining amount
* $25 \text{ g} - 6 \text{ g} = 19 \text{ g}$
Answer for (b): 19 g of $H_2$.
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Problem 2: Magnesium and Hydrochloric Acid Reaction
Reaction: $Mg + 2 HCl \rightarrow MgCl_2 + H_2$
Given: 50 g of $Mg$ and 75 g of $HCl$.
Goal: Find the volume of $H_2$ gas produced at STP.
We need to find the limiting reagent again by seeing which reactant produces less hydrogen gas. At STP (Standard Temperature and Pressure), 1 mole of any gas equals 22.4 Liters.
1. Calculate potential $H_2$ from Magnesium ($Mg$):
* Molar mass of $Mg$ = 24.3 g/mol.
* Ratio: 1 mol $Mg$ makes 1 mol $H_2$.
* Calculation:
$$50 \text{ g } Mg \times \frac{1 \text{ mol } Mg}{24.3 \text{ g } Mg} \times \frac{1 \text{ mol } H_2}{1 \text{ mol } Mg} \times \frac{22.4 \text{ L } H_2}{1 \text{ mol } H_2} \approx 46.1 \text{ L } H_2$$
2. Calculate potential $H_2$ from Hydrochloric Acid ($HCl$):
* Molar mass of $HCl$ = $1 + 35.5 = 36.5$ g/mol.
* Ratio: 2 mol $HCl$ makes 1 mol $H_2$.
* Calculation:
$$75 \text{ g } HCl \times \frac{1 \text{ mol } HCl}{36.5 \text{ g } HCl} \times \frac{1 \text{ mol } H_2}{2 \text{ mol } HCl} \times \frac{22.4 \text{ L } H_2}{1 \text{ mol } H_2} \approx 23.0 \text{ L } H_2$$
3. Compare:
* Magnesium could make 46 L.
* HCl could make 23 L.
* The smaller amount is the correct answer because the HCl runs out first.
Answer: 23 L of $H_2$.
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Problem 3: Zinc and Sulfur Reaction
Reaction: $Zn + S \rightarrow ZnS$
Given: 25 g of Zinc ($Zn$) and 30 g of Sulfur ($S$).
Goal: Identify the limiting reagent.
We compare how much product ($ZnS$) each reactant can make.
1. Calculate potential $ZnS$ from Zinc ($Zn$):
* Molar mass of $Zn$ = 65.4 g/mol.
* Molar mass of $ZnS$ = $65.4 + 32 = 97.4$ g/mol (using 97.48 as per sheet).
* Ratio: 1 mol $Zn$ makes 1 mol $ZnS$.
* Calculation:
$$25 \text{ g } Zn \times \frac{1 \text{ mol } Zn}{65.4 \text{ g } Zn} \times \frac{1 \text{ mol } ZnS}{1 \text{ mol } Zn} \times \frac{97.48 \text{ g } ZnS}{1 \text{ mol } ZnS} \approx 37.3 \text{ g } ZnS$$
2. Calculate potential $ZnS$ from Sulfur ($S$):
* Molar mass of $S$ = 32 g/mol.
* Ratio: 1 mol $S$ makes 1 mol $ZnS$.
* Calculation:
$$30 \text{ g } S \times \frac{1 \text{ mol } S}{32 \text{ g } S} \times \frac{1 \text{ mol } ZnS}{1 \text{ mol } S} \times \frac{97.48 \text{ g } ZnS}{1 \text{ mol } ZnS} \approx 91.4 \text{ g } ZnS$$
3. Compare:
* Zinc produces 37.3 g of product.
* Sulfur produces 91.4 g of product.
* Since Zinc produces the smaller amount, it limits the reaction.
Answer: Zn is the limiting reagent.
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Final Answer:
1a) 34 g NH₃
1b) 19 g H₂ remains
2) 23 L H₂
3) Zn is the limiting reagent
Parent Tip: Review the logic above to help your child master the concept of limiting reagent worksheet answers.
