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 problem, we’ll go through each part of the question step-by-step:
Part (a): Calculating Moles of 2-chloro-2-methylbutane Used
- Volume of 2-chloro-2-methylbutane used: (0.80 \, \text{mL})
- Density of 2-chloro-2-methylbutane: (0.866 \, \text{g/mL})
- Molar mass of 2-chloro-2-methylbutane: (106.58 \, \text{g/mol})
First, we calculate the mass of 2-chloro-2-methylbutane:
[ \text{Mass} = \text{Volume} \times \text{Density} = 0.80 \, \text{mL} \times 0.866 \, \text{g/mL} = 0.6928 \, \text{g} ]
Now, convert the mass to moles:
[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.6928 \, \text{g}}{106.58 \, \text{g/mol}} = 0.00650 \, \text{mol} ]
Part (b): Moles of HCl Produced
The reaction of 2-chloro-2-methylbutane in hydrolysis typically produces one mole of HCl per mole of 2-chloro-2-methylbutane. Thus, the moles of HCl produced would be equal to the moles of 2-chloro-2-methylbutane:
[ \text{Moles of HCl} = 0.00650 \, \text{mol} ]
Part (c): Volume of 0.100M NaOH Needed to Neutralize HCl
To neutralize the HCl produced, we need an equivalent amount of NaOH since the reaction is ( \text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O} ). Thus:
[ \text{Volume of NaOH} = \frac{\text{Moles of HCl}}{\text{Concentration of NaOH}} = \frac{0.00650 \, \text{mol}}{0.100 \, \text{M/L}} = 0.065 \, \text{L} = 65 \, \text{mL} ]
Part (d): Volume of 0.100M NaOH for 80% Completion
For 80% completion, we calculate 80% of the total moles of HCl expected and then calculate the volume of NaOH needed for that:
[ \text{80% Moles of HCl} = 0.80 \times 0.00650 \, \text{mol} = 0.00520 \, \text{mol} ]
[ \text{Volume of NaOH for 80%} = \frac{0.00520 \, \text{mol}}{0.100 \, \text{M/L}} = 0.052 \, \text{L} = 52 \, \text{mL} ]
Summary
- (a) Moles of 2-chloro-2-methylbutane used: (0.00650 \, \text{mol})
- (b) Moles of HCl produced: (0.00650 \, \text{mol})
- (c) mL of 0.100M NaOH required to neutralize: (65 \, \text{mL})
- (d) mL of 0.100M NaOH for 80% completion: (52 \, \text{mL})
These calculations help us understand the stoichiometry and reaction progression in typical organic chemistry hydrolysis reactions, emphasizing the importance of precise measurements and conversions in chemical experiments.