/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Sodium stearate \(\left(\mathrm{... [FREE SOLUTION] | 91影视

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Sodium stearate \(\left(\mathrm{C}_{18} \mathrm{H}_{35} \mathrm{O}_{2} \mathrm{Na}\right)\) is the most common soap. Assume that the stearate anion undergoes aerobic decomposition in the following manner: $$ \begin{aligned} \mathrm{C}_{18} \mathrm{H}_{35} \mathrm{O}_{2}^{-}(a q)+26 \mathrm{O}_{2}(a q) & \longrightarrow \\ & 17 \mathrm{CO}_{2}(a q)+17 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ What is the total mass of \(\mathrm{O}_{2}\) required to biodegrade \(3.0 \mathrm{~g}\) of this substance?

Short Answer

Expert verified
The total mass of O鈧 required to biodegrade 3.0 g of sodium stearate is \(8.64 \: \text{g}\).

Step by step solution

01

Find the moles of sodium stearate

First, we need to find the molar mass of sodium stearate (C鈧佲倛H鈧冣倕O鈧侼a). To do this, we add the molar masses of each element in the compound: Molar mass of C鈧佲倛H鈧冣倕O鈧侼a = 18脳(12.01 amu) + 35脳(1.01 amu) + 2脳(16.00 amu) + 1脳(22.99 amu) = 287.46 g/mol Now, we can calculate the moles of sodium stearate: moles of C鈧佲倛H鈧冣倕O鈧侼a = (mass of sodium stearate)/(molar mass of sodium stearate) moles of C鈧佲倛H鈧冣倕O鈧侼a = (3.0 g)/(287.46 g/mol) = 0.0104 mol
02

Balance the given reaction

The given reaction is already balanced, so we can use the stoichiometric coefficients to find the moles of O鈧 required to react with sodium stearate: C鈧佲倛H鈧冣倕O鈧傗伝 + 26 O鈧 鈫 17 CO鈧 + 17 H鈧侽 + HCO鈧冣伝 According to the balanced equation, 26 moles of O鈧 are needed for every 1 mole of sodium stearate.
03

Calculate the moles of O鈧 required

Using the stoichiometric coefficients from Step 2, we can calculate the moles of O鈧 needed: moles of O鈧 = (moles of C鈧佲倛H鈧冣倕O鈧侼a) 脳 (26 moles O鈧 / 1 mole C鈧佲倛H鈧冣倕O鈧侼a) moles of O鈧 = 0.0104 mol C鈧佲倛H鈧冣倕O鈧侼a 脳 (26 moles O鈧/1 mole C鈧佲倛H鈧冣倕O鈧侼a) = 0.270 mol O鈧
04

Calculate the mass of O鈧 required

Now, we can find the mass of O鈧 required by multiplying the moles of O鈧 by the molar mass of O鈧: mass of O鈧 = moles of O鈧 脳 molar mass of O鈧 mass of O鈧 = 0.270 mol 脳 32.00 g/mol = 8.64 g So, 8.64 g of O鈧 is required to biodegrade 3.0 g of sodium stearate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Stearate
Sodium stearate is a common ingredient found in soap, and it serves as a key component due to its ability to emulsify oils and dirt, making them washable with water. Its chemical formula is represented as \(\text{C}_{18}\text{H}_{35}\text{O}_{2}\text{Na}\), where each element plays a critical role in its function. It consists of a stearate anion, \(\text{C}_{18}\text{H}_{35}\text{O}_{2}^{-}\), and a sodium cation, \(\text{Na}^{+}\).

The structure is derived from stearic acid, a long-chain fatty acid, reacting with sodium hydroxide to form the salt known as sodium stearate. This compound is crucial for the cleansing action because it forms micelles in water.

Micelles trap the oil-based dirt on surfaces using their hydrophobic tails while their hydrophilic heads interact with the water, thereby enabling the dirt to be washed away during rinsing.
Molar Mass
Molar mass is an important concept in stoichiometry and is used for converting grams of a substance into moles, a standard scientific unit for measuring the amount of substance. It acts like a "bridge" in calculations, allowing us to switch between the mass of a compound and the number of molecules or atoms it contains.

To find the molar mass of sodium stearate, \(\text{C}_{18}\text{H}_{35}\text{O}_{2}\text{Na}\), we sum up the atomic masses of its constituent atoms. Specifically, we calculate it as follows:
  • Carbon (C) = 18 atoms 脳 12.01 g/mol
  • Hydrogen (H) = 35 atoms 脳 1.01 g/mol
  • Oxygen (O) = 2 atoms 脳 16.00 g/mol
  • Sodium (Na) = 1 atom 脳 22.99 g/mol
Adding these values gives us the molar mass of 287.46 g/mol. This value is essential when calculating the amount of reactants and products during a chemical process.
Decomposition Reaction
Decomposition reactions involve a single compound breaking down into two or more products. These reactions are important in understanding various chemical processes, especially when analyzing the byproducts and end results of a reaction involving complex molecules.

In this context with sodium stearate, the decomposition refers to the breakdown of \(\text{C}_{18}\text{H}_{35}\text{O}_{2}^{-}\) (stearate anion) when exposed to oxygen, forming carbon dioxide \(\text{(CO}_2)\), water \(\text{(H}_2\text{O})\), and bicarbonate ion \(\text{(HCO}_{3}^{-})\).

The balance of the chemical equation reflects the stoichiometry:
  • 1 mole of \(\text{C}_{18}\text{H}_{35}\text{O}_{2}^{-}\) breaks down with 26 moles of \(\text{O}_{2}\).
  • Producing 17 moles of \(\text{CO}_{2}\), 17 moles of water \(\text{H}_2\text{O}\), and 1 mole of \(\text{HCO}_{3}^{-}\).
This decomposition allows us to determine the exact amounts of each product formed from the initial reactants.
Aerobic Decomposition
Aerobic decomposition is a type of decomposition reaction where oxygen is actively involved. This kind of reaction occurs in environments rich in oxygen and contrasts with anaerobic decomposition that occurs in the absence of oxygen.

In the reaction with sodium stearate, the term "aerobic" highlights the consumption of oxygen during the process. As seen in the equation provided, \(\text{O}_{2}\) is a crucial component that allows the stearate ion to break down efficiently. The presence of oxygen facilitates complete decomposition, creating carbon dioxide and water.

Understanding the mechanics behind aerobic processes is vital, especially in environmental contexts where waste decomposition needs to occur swiftly and effectively, contributing to reducing atmosphere pollutants. Aerobic decomposition is generally faster and more energy efficient, often resulting in fewer odors compared to its anaerobic counterpart due to the absence of partially decomposed products.

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