1. Stoichiometry: Balancing Chemical Equations
Problem 1
Balance the following chemical equation:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
Answer:
To balance the equation, we need to ensure that the number of atoms for each element is the same on both sides.
- From propane (C₃H₈):
- 3 carbon (C) atoms → 3 CO₂
- 8 hydrogen (H) atoms → 4 H₂O
- Oxygen (O) on the product side:
- 3 CO₂ → 3 × 2 = 6 O
- 4 H₂O → 4 × 1 = 4 O
- Total O = 6 + 4 = 10 O
On the reactant side, we have:
- C₃H₈ + O₂ → 1 C₃H₈ + 5 O₂ (since each O₂ contributes 2 O atoms)
The balanced equation is:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
Problem 2
Given 2 moles of \(\text{Na}_2\text{S}\), how many grams of \(\text{S}\) can be produced?
Answer:
The balanced chemical reaction for the formation of sodium sulfide (\(\text{Na}_2\text{S}\)) can be:
\[ 2 \text{Na} + \text{S} \rightarrow \text{Na}_2\text{S} \]
From the equation:
- 1 mole of sulfur (\(S\)) produces 1 mole of sodium sulfide.
- Therefore, 2 moles of \(\text{Na}_2\text{S}\) require 2 moles of \(S\).
To find grams, we use the molar mass of sulfur (approximately 32.07 g/mol):
\[ 2 \text{ mol} \times 32.07 \text{ g/mol} = 64.14 \text{ g} \]
Thus, 64.14 grams of sulfur can be produced.
2. Thermodynamics: Enthalpy Calculations
Problem 3
Calculate the enthalpy change (\(\Delta H\)) for the reaction:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \]
Given that the enthalpy of formation of \(\text{H}_2\text{O}\) is -285.8 kJ/mol.
Answer:
The enthalpy change for the formation of water from hydrogen and oxygen is:
\[ \Delta H = \text{Enthalpy of products} - \text{Enthalpy of reactants} \]
Since the enthalpy of formation for \(\text{H}_2\) and \(\text{O}_2\) is zero (elements in their standard state), we have:
\[ \Delta H = -285.8 \text{ kJ/mol} - (0 + 0) = -285.8 \text{ kJ/mol} \]
Thus, the enthalpy change for the reaction is -285.8 kJ/mol.
Problem 4
If 100 g of \(\text{H}_2\) reacts with excess \(\text{O}_2\), how much energy is released? (Given \(\Delta H = -285.8 \text{ kJ/mol}\))
Answer:
First, we calculate the number of moles of \(\text{H}_2\):
- Molar mass of \(\text{H}_2 = 2 \text{ g/mol}\):
\[ \text{Moles of } \text{H}_2 = \frac{100 \text{ g}}{2 \text{ g/mol}} = 50 \text{ moles} \]
The reaction produces \(\text{H}_2\text{O}\), so:
\[ \text{Energy released} = \text{moles of } \text{H}_2 \times \Delta H \]
\[ = 50 \text{ moles} \times (-285.8 \text{ kJ/mol}) = -14290 \text{ kJ} \]
Thus, 14,290 kJ of energy is released.
3. Chemical Reactions: Types and Classifications
Problem 5
Identify the type of reaction:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
Answer:
This reaction is a decomposition reaction. It involves the breakdown of calcium carbonate (\(\text{CaCO}_3\)) into calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)).
Problem 6
Classify the following reaction:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]
Answer:
This is a double displacement reaction (also known as a double replacement reaction). In this type of reaction, two compounds exchange ions to form two new compounds.
4. Acids and Bases: pH Calculations
Problem 7
Calculate the pH of a solution with a hydrogen ion concentration of \(1 \times 10^{-4} \text{ M}\).
Answer:
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the given concentration:
\[ \text{pH} = -\log(1 \times 10^{-4}) = 4 \]
Thus, the pH of the solution is 4.
Problem 8
If the pH of a solution is 9, what is the concentration of hydroxide ions \([\text{OH}^-]\)?
Answer:
We can use the relationship between \([\text{H}^+]\) and \([\text{OH}^-]\) in water:
\[ \text{pH} + \text{pOH} = 14 \]
First, calculate the pOH:
\[ \text{pOH} = 14 - 9 = 5 \]
Now, find the hydroxide ion concentration:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5} \text{ M} \]
Thus, the concentration of hydroxide ions is \(1 \times 10^{-5} \text{ M}\).
5. Gas Laws: Applications and Calculations
Problem 9
Using the Ideal Gas Law, calculate the volume of 2 moles of gas at 25°C and 1 atm pressure.
Answer:
The Ideal Gas Law is expressed as:
\[ PV = nRT \]
Where:
- \(P\) = pressure (1 atm)
- \(V\) = volume (unknown)
- \(n\) = number of moles (2 moles)
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (25°C = 298 K)
Now, substituting the values:
\[ (1 \text{ atm}) \cdot V = (2 \text{ moles}) \cdot (0.0821 \text{ L·atm/(K·mol)}) \cdot (298 \text{ K}) \]
\[ V = \frac{(2)(0.0821)(298)}{1} = 49.47 \text{ L} \]
Thus, the volume of the gas is approximately 49.47 liters.
Problem 10
If the pressure of a gas is halved while the temperature remains constant, what happens to the volume of the gas?
Answer:
According to Boyle’s Law, which states that pressure and volume are inversely related when temperature is constant:
\[ P_1V_1 = P_2V_2 \]
If \(P_2 = \frac{1}{2}P_1\), then:
\[ P_1V_1 = \frac{1}{2}P_1V_2 \]
Dividing both sides by \(
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