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Chapter 4: Problem 17

Hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) burns in air \(\left(\mathrm{O}_{2}\right)\) to give \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write a balanced equation for the reaction. (b) If \(215 \mathrm{g}\) of \(\mathrm{C}_{6} \mathrm{H}_{14}\) is mixed with \(215 \mathrm{g}\) of \(\mathrm{O}_{2}\), what masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are produced in the reaction? (c) What mass of the excess reactant remains after the hexane has been burned? (d) Set up an amounts table for this problem.

Short Answer

Expert verified
(a) Balanced equation: \(2\mathrm{C}_6\mathrm{H}_{14} + 19\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 14\mathrm{H}_2\mathrm{O}\). (b) \(186.73\) g \(\mathrm{CO}_2\) and \(89.13\) g \(\mathrm{H}_2\mathrm{O}\) are produced. (c) \(184.38\) g of hexane remains.

Step by step solution

01

Write the Unbalanced Equation

First, write the chemical equation for the combustion of hexane. Hexane (\(\mathrm{C}_6\mathrm{H}_{14}\)) reacts with oxygen (\(\mathrm{O}_2\)) to produce carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The unbalanced equation is: \[\mathrm{C}_6\mathrm{H}_{14} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\]
02

Balance the Chemical Equation

To balance the equation, ensure the number of atoms of each element is equal on both sides. Balanced equation: \[2\mathrm{C}_6\mathrm{H}_{14} + 19\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 14\mathrm{H}_2\mathrm{O}\]
03

Calculate Molar Masses

Calculate molar masses: \(\mathrm{C}_6\mathrm{H}_{14}\) is 86 g/mol, \(\mathrm{O}_2\) is 32 g/mol, \(\mathrm{CO}_2\) is 44 g/mol, and \(\mathrm{H}_2\mathrm{O}\) is 18 g/mol. This will be used to convert grams to moles and vice versa.
04

Determine Moles of Reactants

Convert given masses to moles. For hexane: \(n = \frac{215}{86} = 2.5\) moles. For oxygen: \(n = \frac{215}{32} \approx 6.72\) moles.
05

Identify the Limiting Reactant

Using the stoichiometry from the balanced equation, hexane requires 9.5 moles of \(\mathrm{O}_2\) per 2.5 moles. Since we only have 6.72 moles of \(\mathrm{O}_2\), \(\mathrm{O}_2\) is the limiting reactant.
06

Calculate Products Using Limiting Reactant

From 6.72 moles of \(\mathrm{O}_2\), use stoichiometry to determine moles of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\). \(6.72 / 19 = 0.3537\) moles of \(\mathrm{C}_6\mathrm{H}_{14}\) react, producing \(0.3537 \times 12 \approx 4.2444\) moles \(\mathrm{CO}_2\) and \(0.3537 \times 14 \approx 4.9518\) moles \(\mathrm{H}_2\mathrm{O}\).
07

Convert Moles of Products to Mass

Convert moles back to mass: \(\mathrm{CO}_2\) mass is \(4.2444 \times 44 = 186.7344\) g, and \(\mathrm{H}_2\mathrm{O}\) mass is \(4.9518 \times 18 = 89.1324\) g.
08

Calculate Remaining Excess Reactant

Since \(\mathrm{O}_2\) is the limiting reactant, determine how much \(\mathrm{C}_6\mathrm{H}_{14}\) remains. Initially 2.5 moles, react 0.3537 moles, leaving \(2.5 - 0.3537 = 2.1463\) moles. Convert to mass: \(2.1463 \times 86 = 184.3818\) g.
09

Set Up the Amounts Table

Create a table to summarize moles and masses before and after reaction: \[\begin{array}{c|c|c|c}\text{Substance} & \text{Initial Moles} & \text{Final Moles} & \text{Mass Remaining (g)} \hline \mathrm{C}_6\mathrm{H}_{14} & 2.5 & 2.1463 & 184.3818 \ \mathrm{O}_2 & 6.72 & 0 & 0 \hline \mathrm{CO}_2 & 0 & 4.2444 & 186.7344 \ \mathrm{H}_2\mathrm{O} & 0 & 4.9518 & 89.1324 \end{array}\]

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Key Concepts

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

Balanced Chemical Equation
Writing a balanced chemical equation involves ensuring that the same number of atoms of each element exists on both sides of the equation. This maintains the law of conservation of mass, which states that matter cannot be created or destroyed. In the combustion of hexane (\(\mathrm{C}_6\mathrm{H}_{14}\)), the reaction with oxygen (\(\mathrm{O}_2\)) produces carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).
First, the unbalanced equation for this reaction is:\[\mathrm{C}_6\mathrm{H}_{14} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\]
To balance the equation, adjust the coefficients to get:\[2\mathrm{C}_6\mathrm{H}_{14} + 19\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 14\mathrm{H}_2\mathrm{O}\]
Now, each element has the same number of atoms on both sides, thus achieving a balanced chemical equation.
Limiting Reactant
The limiting reactant is the substance that is entirely consumed in a chemical reaction and limits the amount of product formed. To determine the limiting reactant, you compare the mole ratios of the reactants available against what is required by the balanced equation.
In the given reaction, 2.5 moles of hexane would theoretically require 9.5 moles of \(\mathrm{O}_2\), based on their stoichiometric ratio (1:9.5).
However, there are only 6.72 moles of \(\mathrm{O}_2\), less than needed. Therefore, \(\mathrm{O}_2\) is limiting, as it will run out first and prevent more hexane from reacting.
Stoichiometry
Stoichiometry is the study of the quantitative relationships among substances as they participate in chemical reactions. It allows us to predict the amounts of reactants and products involved in a chemical reaction based on the balanced equation.
To find the products formed, use stoichiometry starting from the limiting reactant. Here, since \(\mathrm{O}_2\) is the limiting reactant, use its mole number to calculate the moles of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) that can be formed.
  • The mole ratio of \(\mathrm{O}_2\) to \(\mathrm{CO}_2\) is 19:12, thus giving about 4.2444 moles of \(\mathrm{CO}_2\).
  • For \(\mathrm{H}_2\mathrm{O}\), the mole ratio is 19:14, giving about 4.9518 moles.
Stoichiometry guides us on using the correct ratios to calculate reactants needed or products formed.
Mole Calculations
Mole calculations are essential in determining how much of a substance participates in a reaction, based on its mass. Using the molecular weight of each substance, we can convert from grams to moles and vice versa.
For hexane, with a molar mass of 86 g/mol, 215 g converts to about 2.5 moles. Similarly, 215 g of \(\mathrm{O}_2\) involves about 6.72 moles using its molar mass of 32 g/mol.
Using these mole calculations:
  • Dig into products, with calculated moles for \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\), converting the moles back to mass.
  • This is done by multiplying moles of product by their molar mass (44 g/mol for \(\mathrm{CO}_2\), 18 g/mol for \(\mathrm{H}_2\mathrm{O}\)).
  • Ultimately providing precise mass values for the products formed.
Mastering mole calculations lets you accurately evaluate chemical reaction dynamics.

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Most popular questions from this chapter

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical las it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers \(1 .\) One way to make the compound is called the "chlorohydrin route." $$ \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Another route is the modern catalytic reaction. $$ \mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} $$ (a) Calculate the \(\%\) atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if \(867 \mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

Make the following conversions. In each case, tell whether the solution is acidic or basic. \(\mathbf{p} \mathbf{H}$$\quad$$\left[\mathbf{H}_{3} \mathbf{O}^{*}\right]\) (a) \(\quad\)______\(\quad 6.7 \times 10^{-10} \mathrm{M}\) (b) \(\quad\)______\(\quad2.2 \times 10^{-6} \mathrm{M}\) (c) 5.25\(\quad\)_____ (d) ______\(\quad2.5 \times 10^{-2} \mathrm{M}\)

Stoichiometry of Reactions in Solution What volume of \(0.109 \mathrm{M} \mathrm{HNO}_{3},\) in milliliters, is required to react completely with \(2.50 \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\) \(2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\) $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) $$

Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen. \(4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq}) $$ (a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced? (b) If you have exactly one metric ton ( 1 metric ton \(=1000 \mathrm{kg})\) of gold-bearing rock, what volume of \(0.075 \mathrm{M} \mathrm{NaCN},\) in liters, do you need to extract the gold if the rock is \(0.019 \%\) gold?

You have \(250 .\) mL of \(0.136 \mathrm{M}\) HCl. Using a volumetric pipet, you take \(25.00 \mathrm{mL}\) of that solution and dilute it to \(100.00 \mathrm{mL}\) in a volumetric flask. Now you take \(10.00 \mathrm{mL}\) of that solution, using a volumetric pipet, and dilute it to \(100.00 \mathrm{mL}\) in a volumetric flask. What is the concentration of hydrochloric acid in the final solution?
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