QuestionThe following information was obtained when carrying out an experiment to determine the Enthalpy $(\Delta \mathrm{H})$ of neutralization reaction between HCl and NaOH . 100 mL of 2.00 M HCl solution was added to 95.00 mL of 2.00 M solution of NaOH . The final temperature reached was $35.40^{\circ} \mathrm{C}$ and the initial temperature at mixing was $22.15^{\circ} \mathrm{C}$ . The density of the mixture was 1.04 $\mathrm{g} / \mathrm{mL}$ and its specific heat capacity was $3.89 \mathrm{~J} / \mathrm{g}-{ }^{\circ} \mathrm{C}$ . How many moles of HCl were used to neutralize the NaOH in the reaction? (A) 0.19 mol (B) 0.20 mol (C) 0.39 mol (D) 95 mol

Studdy Solution

STEP 1

1. The reaction between HCl and NaOH is a 1:1 stoichiometric reaction.
2. The concentration of HCl is $2.00 \, \text{M}$ .
3. The volume of HCl used is $100 \, \text{mL}$ .

STEP 2

1. Convert the volume of HCl from milliliters to liters.
2. Use the molarity formula to calculate the moles of HCl.
3. Verify the stoichiometry of the reaction.

STEP 3

Convert the volume of HCl from milliliters to liters:
$100 \, \text{mL} = 0.100 \, \text{L}$

STEP 4

Use the molarity formula to calculate the moles of HCl:
The formula for moles is:
$\text{Moles} = \text{Molarity} \times \text{Volume (in Liters)}$
Substitute the given values:
$\text{Moles of HCl} = 2.00 \, \text{M} \times 0.100 \, \text{L}$
$= 0.200 \, \text{mol}$

STEP 5

Verify the stoichiometry of the reaction:
Since the reaction between HCl and NaOH is 1:1, the moles of HCl used is equal to the moles of NaOH neutralized.
The number of moles of HCl used to neutralize the NaOH is:
$\boxed{0.20 \, \text{mol}}$

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