To find out how much heat is released when 74.3 mL of 1.750 M HCl is reacted with 74.3 mL of 1.750 M NaOH, we can follow these steps:
### Step 1: Calculate the moles of HCl and NaOH
The molarity (M) of a solution tells us how many moles of solute are in 1 liter (L) of solution. Therefore, we can calculate the number of moles of HCl and NaOH in the given volumes of their solutions by using the formula:
\[ \text{moles} = \text{Molarity} \times \text{Volume in liters} \]
Given:
- Volume of HCl = 74.3 mL = 0.0743 L (since 1000 mL = 1 L)
- Molarity of HCl = 1.750 M
- Volume of NaOH = 74.3 mL = 0.0743 L
- Molarity of NaOH = 1.750 M
### Step 2: Calculate the moles of reactants
\[ \text{Moles of HCl} = 1.750 \, \text{M} \times 0.0743 \, \text{L} \]
\[ \text{Moles of NaOH} = 1.750 \, \text{M} \times 0.0743 \, \text{L} \]
### Step 3: Determine the limiting reactant
Since both HCl and NaOH are present in equal molar quantities and the reaction between HCl and NaOH is 1:1, they will completely react with each other. Thus, either can be considered the limiting reactant, and all of it will be used up.
### Step 4: Calculate the heat released
The heat of neutralization (ΔH) is given as –56.2 kJ/mol. This is the amount of heat released per mole of HCl (or NaOH) reacted.
Using the moles calculated and the heat of neutralization:
\[ \text{Total heat released} = \text{moles of HCl (or NaOH)} \times \text{Heat of neutralization per mole} \]
Let's compute these values.
The total heat released when 74.3 mL of 1.750 M HCl is reacted with 74.3 mL of 1.750 M NaOH is approximately -7.31 kJ. This value matches option C, which is 7.31 kJ.
The process involves calculating the moles of reactants (which are equal for HCl and NaOH due to their equal molarity and volume), and then using the heat of neutralization to determine the total amount of heat released. Since the reaction is stoichiometric with a 1:1 ratio and both reactants are in equal amounts, all of the reactants are completely used, ensuring that the calculated heat is accurate. Thus, the correct answer is:
**C) 7.31 kJ**
