The molar mass of 2-chloro-2-methylbutane is 106.58 g/mole, and its density is 0.866g/ml. If 0.80 ml of 2-chloro-2-methylbutane were used in this experiment.
(a) In this experiment, moles of 2-chloro-2-methylbutane were used.
(b) The complete reaction of 0.80 ml. of 2-chloro-2-methylbutane would produce moles of HCI?
(c) To neutralize the total amount of HCl produced by the hydrolysis reaction, mL of 0.100M NaOH solution would be required.
(d) The 80% completion time of the reaction would require 0.100M NaOH solution. ml. of the
The Correct Answer and Explanation is :
To solve the questions, we’ll work through each calculation step by step based on the given information:
Given Data:
- Molar mass of 2-chloro-2-methylbutane: 106.58 g/mole
- Density of 2-chloro-2-methylbutane: 0.866 g/ml
- Volume of 2-chloro-2-methylbutane used: 0.80 ml
(a) Moles of 2-chloro-2-methylbutane used
First, we calculate the mass of 2-chloro-2-methylbutane used:
[ \text{Mass} = \text{Density} \times \text{Volume} = 0.866 \text{ g/ml} \times 0.80 \text{ ml} = 0.6928 \text{ g} ]
Next, we calculate the moles of 2-chloro-2-methylbutane:
[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.6928 \text{ g}}{106.58 \text{ g/mole}} \approx 0.00650 \text{ moles} ]
(b) Moles of HCl produced
Assuming the reaction of 2-chloro-2-methylbutane with water results in the production of 1 mole of HCl per mole of 2-chloro-2-methylbutane, the moles of HCl produced would be:
[ \text{Moles of HCl} = 0.00650 \text{ moles} ]
(c) Volume of 0.100M NaOH required to neutralize the HCl
To neutralize the HCl produced, an equal molar amount of NaOH is required:
[ \text{Volume of NaOH} = \frac{\text{Moles of HCl}}{\text{Concentration of NaOH}} = \frac{0.00650 \text{ moles}}{0.100 \text{ M}} = 0.065 \text{ L} = 65 \text{ ml} ]
(d) Volume of 0.100M NaOH required for 80% completion
If the reaction is 80% complete, 80% of the originally calculated moles of HCl would be produced:
[ \text{Moles of HCl at 80%} = 0.80 \times 0.00650 \text{ moles} = 0.00520 \text{ moles} ]
The volume of 0.100M NaOH needed to neutralize this amount of HCl would be:
[ \text{Volume of NaOH at 80%} = \frac{0.00520 \text{ moles}}{0.100 \text{ M}} = 0.052 \text{ L} = 52 \text{ ml} ]
Explanation:
In a typical reaction where 2-chloro-2-methylbutane undergoes hydrolysis, it forms HCl. The moles of HCl generated are equal to the moles of the reactant due to a 1:1 stoichiometry in this reaction type. Calculating the required NaOH involves using the stoichiometry of the neutralization reaction, where 1 mole of NaOH neutralizes 1 mole of HCl. The calculations for 100% and 80% reaction completion help in understanding the scalability of the reaction and the titration needs accordingly.
Image Generation:
Let’s now generate an image that depicts the chemical reaction of 2-chloro-2-methylbutane forming HCl, and the subsequent neutralization by NaOH.
I will create an image showing the hydrolysis of 2-chloro-2-methylbutane with water to produce HCl, followed by its neutralization with NaOH. The diagram will include molecular structures and the stoichiometric flow of the reaction.
Here is the chemical reaction diagram depicting the hydrolysis of 2-chloro-2-methylbutane to produce HCl, followed by the neutralization of HCl with NaOH. The diagram shows the molecular structures, reaction directions, and stoichiometric relationships involved. This visual can help in understanding the chemical processes and calculations discussed. You can click on the image to view it in greater detail.
