Mixed Stoichiometry Practice Worksheet: Complete with ease ... - Free Printable

Problem Analysis:

The task involves solving stoichiometry problems, which require balancing chemical equations and using the balanced equations to perform mole-to-mole, mass-to-mass, or other conversions. Let's solve each part step by step.

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Problem 1: Potassium chlorate decomposes into potassium chloride and oxygen gas.

#### Step 1: Write the unbalanced equation.
Potassium chlorate ($\text{KClO}_3$) decomposes into potassium chloride ($\text{KCl}$) and oxygen gas ($\text{O}_2$):
$$
\text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2
$$

#### Step 2: Balance the equation.
- Start with potassium (K): There is 1 K on both sides.
- Next, chlorine (Cl): There is 1 Cl on both sides.
- Finally, oxygen (O): There are 3 O atoms in $\text{KClO}_3$, but only 2 O atoms in $\text{O}_2$. To balance oxygen, we need a coefficient of 3 for $\text{O}_2$ and 2 for $\text{KClO}_3$:
$$
2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2
$$

#### Balanced Equation:
$$
\boxed{2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2}
$$

#### Step 3: Calculate moles of oxygen produced when 3.0 moles of $\text{KClO}_3$ decompose.
From the balanced equation:
$$
2\text{KClO}_3 \rightarrow 3\text{O}_2
$$
This means that 2 moles of $\text{KClO}_3$ produce 3 moles of $\text{O}_2$. Therefore, the mole ratio is:
$$
\frac{\text{moles of } \text{O}_2}{\text{moles of } \text{KClO}_3} = \frac{3}{2}
$$

If 3.0 moles of $\text{KClO}_3$ decompose:
$$
\text{Moles of } \text{O}_2 = 3.0 \times \frac{3}{2} = 4.5 \text{ moles}
$$

#### Answer for Problem 2:
$$
\boxed{4.5 \text{ moles}}
$$

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Problem 3: Butane ($\text{C}_4\text{H}_{10}$) undergoes combustion.

#### Step 1: Write the unbalanced equation.
Butane ($\text{C}_4\text{H}_{10}$) reacts with oxygen ($\text{O}_2$) to produce carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$):
$$
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

#### Step 2: Balance the equation.
- Carbon (C): There are 4 C atoms in $\text{C}_4\text{H}_{10}$, so we need 4 $\text{CO}_2$ molecules:
$$
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + \text{H}_2\text{O}
$$
- Hydrogen (H): There are 10 H atoms in $\text{C}_4\text{H}_{10}$, so we need 5 $\text{H}_2\text{O}$ molecules:
$$
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}
$$
- Oxygen (O): On the right side, there are $4 \times 2 = 8$ O atoms from $\text{CO}_2$ and $5 \times 1 = 5$ O atoms from $\text{H}_2\text{O}$, totaling 13 O atoms. Since each $\text{O}_2$ molecule provides 2 O atoms, we need $\frac{13}{2} = 6.5$ moles of $\text{O}_2$. To avoid fractions, multiply the entire equation by 2:
$$
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
$$

#### Balanced Equation:
$$
\boxed{2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}}
$$

#### Step 3: Calculate grams of $\text{CO}_2$ produced when 88 g of $\text{O}_2$ react.
First, find the molar mass of $\text{O}_2$:
$$
\text{Molar mass of } \text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}
$$
Convert 88 g of $\text{O}_2$ to moles:
$$
\text{Moles of } \text{O}_2 = \frac{88 \text{ g}}{32.00 \text{ g/mol}} = 2.75 \text{ moles}
$$

From the balanced equation:
$$
13 \text{ moles of } \text{O}_2 \rightarrow 8 \text{ moles of } \text{CO}_2
$$
So, the mole ratio is:
$$
\frac{\text{moles of } \text{CO}_2}{\text{moles of } \text{O}_2} = \frac{8}{13}
$$
Calculate moles of $\text{CO}_2$:
$$
\text{Moles of } \text{CO}_2 = 2.75 \times \frac{8}{13} = 1.6923 \text{ moles}
$$

Next, find the molar mass of $\text{CO}_2$:
$$
\text{Molar mass of } \text{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \text{ g/mol}
$$
Convert moles of $\text{CO}_2$ to grams:
$$
\text{Mass of } \text{CO}_2 = 1.6923 \text{ moles} \times 44.01 \text{ g/mol} = 74.5 \text{ g}
$$

#### Answer for Problem 4:
$$
\boxed{74.5 \text{ g}}
$$

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Problem 5: Water decomposes into hydrogen gas and oxygen gas by electrolysis.

#### Step 1: Write the unbalanced equation.
Water ($\text{H}_2\text{O}$) decomposes into hydrogen gas ($\text{H}_2$) and oxygen gas ($\text{O}_2$):
$$
\text{H}_2\text{O} \rightarrow \text{H}_2 + \text{O}_2
$$

#### Step 2: Balance the equation.
- Hydrogen (H): There are 2 H atoms in $\text{H}_2\text{O}$, so we need 1 $\text{H}_2$ molecule.
- Oxygen (O): There are 2 O atoms in $\text{H}_2\text{O}$, so we need $\frac{1}{2}$ $\text{O}_2$ molecule. To avoid fractions, multiply the entire equation by 2:
$$
2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2
$$

#### Balanced Equation:
$$
\boxed{2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2}
$$

#### Step 3: Calculate grams of hydrogen produced when 6.0 moles of $\text{O}_2$ are produced.
From the balanced equation:
$$
2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2
$$
This means that 1 mole of $\text{O}_2$ produces 2 moles of $\text{H}_2$. Therefore, the mole ratio is:
$$
\frac{\text{moles of } \text{H}_2}{\text{moles of } \text{O}_2} = 2
$$

If 6.0 moles of $\text{O}_2$ are produced:
$$
\text{Moles of } \text{H}_2 = 6.0 \times 2 = 12.0 \text{ moles}
$$

Next, find the molar mass of $\text{H}_2$:
$$
\text{Molar mass of } \text{H}_2 = 2 \times 1.01 = 2.02 \text{ g/mol}
$$
Convert moles of $\text{H}_2$ to grams:
$$
\text{Mass of } \text{H}_2 = 12.0 \text{ moles} \times 2.02 \text{ g/mol} = 24.24 \text{ g}
$$

#### Answer for Problem 6:
$$
\boxed{24.2 \text{ g}}
$$

#### Step 4: Calculate grams of water required to produce 9.00 grams of hydrogen.
First, find the molar mass of $\text{H}_2$:
$$
\text{Molar mass of } \text{H}_2 = 2.02 \text{ g/mol}
$$
Convert 9.00 g of $\text{H}_2$ to moles:
$$
\text{Moles of } \text{H}_2 = \frac{9.00 \text{ g}}{2.02 \text{ g/mol}} = 4.455 \text{ moles}
$$

From the balanced equation:
$$
2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2
$$
This means that 2 moles of $\text{H}_2\text{O}$ produce 2 moles of $\text{H}_2$. Therefore, the mole ratio is:
$$
\frac{\text{moles of } \text{H}_2\text{O}}{\text{moles of } \text{H}_2} = 1
$$
So, moles of $\text{H}_2\text{O}$ required:
$$
\text{Moles of } \text{H}_2\text{O} = 4.455 \text{ moles}
$$

Next, find the molar mass of $\text{H}_2\text{O}$:
$$
\text{Molar mass of } \text{H}_2\text{O} = 2 \times 1.01 + 16.00 = 18.02 \text{ g/mol}
$$
Convert moles of $\text{H}_2\text{O}$ to grams:
$$
\text{Mass of } \text{H}_2\text{O} = 4.455 \text{ moles} \times 18.02 \text{ g/mol} = 80.3 \text{ g}
$$

#### Answer for Problem 7:
$$
\boxed{80.3 \text{ g}}
$$

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Final Answers:

1. Balanced Equation: $\boxed{2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2}$
2. Moles of $\text{O}_2$: $\boxed{4.5 \text{ moles}}$
3. Balanced Equation: $\boxed{2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}}$
4. Grams of $\text{CO}_2$: $\boxed{74.5 \text{ g}}$
5. Balanced Equation: $\boxed{2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2}$
6. Grams of $\text{H}_2$: $\boxed{24.2 \text{ g}}$
7. Grams of $\text{H}_2\text{O}$: $\boxed{80.3 \text{ g}}$