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Chapter 3: Problem 130

Hexamethylenediamine $\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)$ is one of the starting materials for the production of nylon. It can be prepared from adipic acid $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)$ by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?

Short Answer

Expert verified
The percent yield for the reaction is 109.41%, which is not possible as it cannot be greater than 100%. There might be errors in the measurements or calculations.

Step by step solution

01

Calculate the number of moles of adipic acid

To find the number of moles of adipic acid, we'll use its molar mass and the given mass: $$ \text{Moles of adipic acid} = \frac{\text{Mass of adipic acid}}{\text{Molar mass of adipic acid}} $$ The molar mass of adipic acid (C6H10O4) is: $$ (6 \times 12.01 + 10 \times 1.01 + 4 \times 16.00)g/mol = 166.14 \,g/mol $$ So, the number of moles of adipic acid is: $$ \text{Moles of adipic acid} = \frac{1.00 \times 10^3 g}{166.14 \,g/mol} = 6.018 \,mol $$
02

Determine the molar ratio of hexamethylenediamine to adipic acid

From the balanced chemical equation, we can see that 1 mole of adipic acid is needed to produce 1 mole of hexamethylenediamine: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l) $$ So, the molar ratio of hexamethylenediamine to adipic acid is 1:1.
03

Calculate the theoretical yield of hexamethylenediamine

Since we have a 1:1 molar ratio, we can use the number of moles of adipic acid to find the theoretical yield of hexamethylenediamine: $$ \text{Moles of hexamethylenediamine} = 6.018 \,mol $$ Now, we will convert moles of hexamethylenediamine to mass using its molar mass: $$ \text{Theoretical yield} = \text{Moles of hexamethylenediamine} \times \text{Molar mass of hexamethylenediamine} $$ The molar mass of hexamethylenediamine (C6H16N2) is: $$ (6 \times 12.01 + 16 \times 1.01 + 2 \times 14.01)g/mol = 116.21 \,g/mol $$ So, the theoretical yield of hexamethylenediamine is: $$ \text{Theoretical yield} = 6.018 \,mol \times 116.21 \,g/mol = 699.26 \,g $$
04

Calculate the percent yield

Finally, we will calculate the percent yield using the actual and theoretical yields: $$ \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% $$ We are given the actual yield as 765 g. Therefore, the percent yield is: $$ \text{Percent yield} = \frac{765 \,g}{699.26 \,g} \times 100\% = 109.41\% $$ However, it's not possible to obtain a yield greater than 100%. This may be due to some error in the measurements or calculations.

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Most popular questions from this chapter

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$ \begin{array}{l}{\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\\ {\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\) . Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

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Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A 10.00 -g mixture of zinc and magnesium produces 0.5171 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.
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