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Chapter 15: Problem 308

When \(10.0 \mathrm{~g}\) of n-propylchloride is allowed to react with excess sodium in the Wurtz reaction, how many grams of hexane would be produced assuming a \(70 \%\) yield?

Short Answer

Expert verified
The amount of hexane produced in the Wurtz reaction with a 70% yield is approximately 3.837 grams.

Step by step solution

01

1. Find the molar mass of n-propylchloride

First, we need to determine the molar mass of n-propylchloride, which has the molecular formula C3H7Cl. The molar masses of each element are: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Chlorine (Cl): 35.45 g/mol Now, we can calculate the molar mass of n-propylchloride: Molar mass of n-propylchloride = (3 x 12.01) + (7 x 1.01) + 35.45 = 78.54 g/mol
02

2. Calculate the moles of n-propylchloride reacted

Given 10.0 g of n-propylchloride, we can find the moles by dividing the mass by the molar mass: Moles of n-propylchloride = (10.0 g) / (78.54 g/mol) = 0.1273 mol
03

3. Use stoichiometry to determine the moles of hexane produced

In the Wurtz reaction, two moles of n-propylchloride react to produce one mole of hexane. Therefore, the ratio between the moles of n-propylchloride and moles of hexane is 2:1. By using this stoichiometric ratio, we can calculate the moles of hexane produced: Moles of hexane = (0.1273 mol of n-propylchloride) * (1 mol of hexane / 2 mol of n-propylchloride) = 0.06365 mol of hexane
04

4. Calculate the molar mass of hexane

Hexane has the molecular formula C6H14. The molar masses of each element are: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol Now, we can calculate the molar mass of hexane: Molar mass of hexane = (6 x 12.01) + (14 x 1.01) = 86.18 g/mol
05

5. Determine the mass of hexane produced

To find the mass of hexane produced, multiply the number of moles of hexane by its molar mass: Mass of hexane = (0.06365 mol of hexane) * (86.18 g/mol) = 5.481 g of hexane
06

6. Adjust for the actual yield of the reaction

The problem states that the yield of the reaction is 70%. To find the actual mass of hexane produced, we need to multiply the theoretical mass of hexane by the actual yield: Mass of hexane (actual) = (5.481 g) * (70%) = 3.837 g So, the amount of hexane produced in this reaction with a 70% yield is approximately 3.837 g.

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Most popular questions from this chapter

An aromatic dibromide \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{Br}_{2}\) reacted with aqueous sodium hydroxide. The product of this reaction had lost only one bromo group to give the product \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{BrO}\). When the dibromide was converted to a Grignard reagent and then hydrolyzed, the product was toluene. Determine the structure of the dibromide.

Write balanced equations for the preparation of each of the following organometallic compounds by two different reactions starting from suitable alky1 halides and inorganic reagents. Specify reaction conditions and solvents. In each case, indicate which method of preparation you would prefer from standpoints of yield, convenience, etc. (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Zn}\) (c) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{4} \mathrm{~Pb}\) (b) \(\mathrm{CH}_{3} \mathrm{MgCl}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHLi}\)

The following projected synthesis for \(\mathrm{n}\) -butane is not very efficient. Why? \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{Br}+2 \mathrm{Na} \rightarrow \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}+2 \mathrm{NaBr}\)

Excess methylmagnesium iodide and \(0.1776 \mathrm{~g}\) of compound A of formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{3}\) Sreact to give \(84.1 \mathrm{cc}\) of methane collected over mercury at \(740 \mathrm{~mm}\) and \(25^{\circ} \mathrm{C}\). How many active hydrogens does compound A possess? Suggest a possible structure for the compound given that the infrared spectrum shows no carbonyl absorption and the n.m.r. spectrum shows only three types of hydrogen with areas in the ratio of \(1: 2: 2\).

Write structures for the products of the following reactions involving Grignard reagents. Show the structures of both the intermediate substances and the substances obtained after hydrolysis with dilute acid. Unless otherwise specified, assume that sufficient Grignard reagent is used to cause those reactions to go to completion which occur readily at room temperatures. (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{MgI}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CMgCl}+\mathrm{CO}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{ClCO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\) (e) \(\mathrm{CH}_{3} \mathrm{Mg} 1+\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\) \((1\) mole \()\) \((1\) mole \()\) 0
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