Limiting-Reagent-Worksheet Worksheet - Free Printable

Problem Analysis and Solution

The worksheet provided involves two chemical reactions, and the tasks are to balance the reactions, determine the limiting reagent, and calculate the amount of product formed. Below is a detailed explanation of each problem.

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#### Problem 1: Reaction Involving Propane (\( \text{C}_3\text{H}_8 \)) and Oxygen (\( \text{O}_2 \))

##### Reaction:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

##### Step 1: Balance the Reaction
To balance the reaction:
- Start with carbon (\( \text{C} \)): There are 3 carbon atoms in \( \text{C}_3\text{H}_8 \), so we need 3 molecules of \( \text{CO}_2 \).
- Next, balance hydrogen (\( \text{H} \)): There are 8 hydrogen atoms in \( \text{C}_3\text{H}_8 \), so we need 4 molecules of \( \text{H}_2\text{O} \) (since each \( \text{H}_2\text{O} \) contains 2 hydrogen atoms).
- Finally, balance oxygen (\( \text{O} \)): Each \( \text{CO}_2 \) molecule has 2 oxygen atoms, and each \( \text{H}_2\text{O} \) molecule has 1 oxygen atom. Therefore, \( 3 \times 2 + 4 \times 1 = 10 \) oxygen atoms are needed on the product side. Since each \( \text{O}_2 \) molecule contains 2 oxygen atoms, we need \( \frac{10}{2} = 5 \) molecules of \( \text{O}_2 \).

The balanced equation is:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

##### Step 2: Determine the Limiting Reagent
Given:
- \( 14.8 \, \text{g} \) of \( \text{C}_3\text{H}_8 \)
- \( 3.44 \, \text{g} \) of \( \text{O}_2 \)

We need to determine which reactant will produce less water (\( \text{H}_2\text{O} \)) and thus is the limiting reagent.

###### Calculate the amount of \( \text{H}_2\text{O} \) produced by \( \text{C}_3\text{H}_8 \):
1. Convert grams of \( \text{C}_3\text{H}_8 \) to moles:
\[
\text{Moles of } \text{C}_3\text{H}_8 = \frac{14.8 \, \text{g}}{44.09 \, \text{g/mol}} = 0.3357 \, \text{mol}
\]
2. Use the stoichiometry of the balanced equation to find the moles of \( \text{H}_2\text{O} \):
\[
\text{Moles of } \text{H}_2\text{O} = 0.3357 \, \text{mol} \, \text{C}_3\text{H}_8 \times \frac{4 \, \text{mol} \, \text{H}_2\text{O}}{1 \, \text{mol} \, \text{C}_3\text{H}_8} = 1.3428 \, \text{mol} \, \text{H}_2\text{O}
\]
3. Convert moles of \( \text{H}_2\text{O} \) to grams:
\[
\text{Mass of } \text{H}_2\text{O} = 1.3428 \, \text{mol} \times 18.02 \, \text{g/mol} = 24.2 \, \text{g}
\]

###### Calculate the amount of \( \text{H}_2\text{O} \) produced by \( \text{O}_2 \):
1. Convert grams of \( \text{O}_2 \) to moles:
\[
\text{Moles of } \text{O}_2 = \frac{3.44 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1075 \, \text{mol}
\]
2. Use the stoichiometry of the balanced equation to find the moles of \( \text{H}_2\text{O} \):
\[
\text{Moles of } \text{H}_2\text{O} = 0.1075 \, \text{mol} \, \text{O}_2 \times \frac{4 \, \text{mol} \, \text{H}_2\text{O}}{5 \, \text{mol} \, \text{O}_2} = 0.0860 \, \text{mol} \, \text{H}_2\text{O}
\]
3. Convert moles of \( \text{H}_2\text{O} \) to grams:
\[
\text{Mass of } \text{H}_2\text{O} = 0.0860 \, \text{mol} \times 18.02 \, \text{g/mol} = 1.55 \, \text{g}
\]

Since \( \text{O}_2 \) produces less \( \text{H}_2\text{O} \) (1.55 g) than \( \text{C}_3\text{H}_8 \) (24.2 g), \( \text{O}_2 \) is the limiting reagent.

##### Step 3: Determine the Amount of \( \text{CO}_2 \) Produced
Using the limiting reagent (\( \text{O}_2 \)):
1. Moles of \( \text{O}_2 \) available: \( 0.1075 \, \text{mol} \)
2. Use the stoichiometry of the balanced equation to find the moles of \( \text{CO}_2 \):
\[
\text{Moles of } \text{CO}_2 = 0.1075 \, \text{mol} \, \text{O}_2 \times \frac{3 \, \text{mol} \, \text{CO}_2}{5 \, \text{mol} \, \text{O}_2} = 0.0645 \, \text{mol} \, \text{CO}_2
\]
3. Convert moles of \( \text{CO}_2 \) to grams:
\[
\text{Mass of } \text{CO}_2 = 0.0645 \, \text{mol} \times 44.01 \, \text{g/mol} = 2.84 \, \text{g}
\]

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#### Problem 2: Reaction Involving Magnesium (\( \text{Mg} \)) and Oxygen (\( \text{O}_2 \))

##### Reaction:
\[ \text{Mg} + \text{O}_2 \rightarrow \text{MgO} \]

##### Step 1: Balance the Reaction
To balance the reaction:
- Each magnesium atom combines with one oxygen atom to form \( \text{MgO} \). However, oxygen exists as \( \text{O}_2 \) in its diatomic form. Therefore, we need 2 magnesium atoms to react with 1 molecule of \( \text{O}_2 \) to produce 2 molecules of \( \text{MgO} \).

The balanced equation is:
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

##### Step 2: Determine the Limiting Reagent
Given:
- \( 2.2 \, \text{g} \) of \( \text{Mg} \)
- \( 4.5 \, \text{L} \) of \( \text{O}_2 \) at STP

We need to determine which reactant will produce less \( \text{MgO} \) and thus is the limiting reagent.

###### Calculate the amount of \( \text{MgO} \) produced by \( \text{Mg} \):
1. Convert grams of \( \text{Mg} \) to moles:
\[
\text{Moles of } \text{Mg} = \frac{2.2 \, \text{g}}{24.31 \, \text{g/mol}} = 0.0905 \, \text{mol}
\]
2. Use the stoichiometry of the balanced equation to find the moles of \( \text{MgO} \):
\[
\text{Moles of } \text{MgO} = 0.0905 \, \text{mol} \, \text{Mg} \times \frac{2 \, \text{mol} \, \text{MgO}}{2 \, \text{mol} \, \text{Mg}} = 0.0905 \, \text{mol} \, \text{MgO}
\]

###### Calculate the amount of \( \text{MgO} \) produced by \( \text{O}_2 \):
1. Convert liters of \( \text{O}_2 \) at STP to moles (1 mole of gas at STP occupies 22.4 L):
\[
\text{Moles of } \text{O}_2 = \frac{4.5 \, \text{L}}{22.4 \, \text{L/mol}} = 0.2009 \, \text{mol}
\]
2. Use the stoichiometry of the balanced equation to find the moles of \( \text{MgO} \):
\[
\text{Moles of } \text{MgO} = 0.2009 \, \text{mol} \, \text{O}_2 \times \frac{2 \, \text{mol} \, \text{MgO}}{1 \, \text{mol} \, \text{O}_2} = 0.4018 \, \text{mol} \, \text{MgO}
\]

Since \( \text{Mg} \) produces less \( \text{MgO} \) (0.0905 mol) than \( \text{O}_2 \) (0.4018 mol), \( \text{Mg} \) is the limiting reagent.

##### Step 3: Determine the Amount of \( \text{MgO} \) Produced
Using the limiting reagent (\( \text{Mg} \)):
1. Moles of \( \text{MgO} \) produced: \( 0.0905 \, \text{mol} \)
2. Convert moles of \( \text{MgO} \) to grams:
\[
\text{Mass of } \text{MgO} = 0.0905 \, \text{mol} \times 40.31 \, \text{g/mol} = 3.64 \, \text{g}
\]

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Final Answers:

1. Limiting Reagent for Problem 1: \( \boxed{\text{O}_2} \)
- Amount of \( \text{CO}_2 \) produced: \( \boxed{2.84 \, \text{g}} \)

2. Limiting Reagent for Problem 2: \( \boxed{\text{Mg}} \)
- Amount of \( \text{MgO} \) produced: \( \boxed{3.64 \, \text{g}} \)