Consider the unbalanced equation for the combustion of butane: $$ \mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Balance the equation and determine how many moles of \(\mathrm{O}_{2}\) are required to react completely with \(4.9 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10}\).

Short Answer

Expert verified
The balanced equation is \(2\mathrm{C}_{4} \mathrm{H}_{10}(g) + 13\mathrm{O}_{2}(g) \longrightarrow 8\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\), and 31.85 moles of \(\mathrm{O}_{2}\) are required to react with 4.9 moles of \(\mathrm{C}_{4} \mathrm{H}_{10}\).

Step by step solution

01

Write the Unbalanced Equation

Set up the combustion reaction for butane: \(\mathrm{C}_{4} \mathrm{H}_{10}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\).
02

Balance Carbon Atoms

Balance the carbon atoms first by ensuring the number of carbon atoms in the products equals the number in the reactants. There are four carbon atoms in butane, so put a coefficient of 4 in front of \(\mathrm{CO}_{2}: \mathrm{C}_{4} \mathrm{H}_{10}(g) + \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\).
03

Balance Hydrogen Atoms

Balance the hydrogen atoms next. Butane has ten hydrogens; therefore put a coefficient of 5 in front of \(\mathrm{H}_2\mathrm{O}\) to get 10 hydrogen atoms in the products: \(\mathrm{C}_{4} \mathrm{H}_{10}(g) + \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(g)\).
04

Balance Oxygen Atoms

Lastly, balance the oxygen atoms. The products now have \(4 \times 2 + 5 \times 1 = 13\) oxygen atoms in total. As oxygen molecules \(\mathrm{O}_2\) contain 2 atoms each, you would need \(\frac{13}{2}\) or 6.5 molecules of \(\mathrm{O}_2\) to balance the oxygen atoms. Write 6.5 in front of the \(\mathrm{O}_2\): \(\mathrm{C}_{4} \mathrm{H}_{10}(g) + 6.5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(g)\).
05

Make Coefficients Whole Numbers

Since we cannot have half a molecule, multiply all coefficients by 2 to make them whole numbers: \(2\mathrm{C}_{4} \mathrm{H}_{10}(g) + 13\mathrm{O}_{2}(g) \longrightarrow 8\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\).
06

Calculate Moles of O2 Needed

We have a balanced equation with a 2:13 ratio between butane and oxygen. Since 4.9 moles of butane (\(\mathrm{C}_{4} \mathrm{H}_{10}\)) are given, to find the moles of \(\mathrm{O}_2\) needed, set up a proportion: \(\frac{4.9\,\text{mol C}_{4} \text{H}_{10}}{2} = \frac{x\,\text{mol O}_{2}}{13}\). Solve for x to get \(x = \frac{4.9 \times 13}{2} = 31.85\text{ mol O}_{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stoichiometry
Stoichiometry is the section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's a foundational concept for understanding how to balance chemical equations and to predict the outcomes of reactions quantitatively. In a balanced chemical equation, the number of atoms for each element must be the same on both sides, respecting the law of conservation of mass. This balance is essential because it reflects the precise proportion of reactants that will combine to form products.
Stoichiometry relies on the mole concept, as it allows chemists to count atoms, molecules, or ions in a given mass of a substance. Utilizing the mole ratio from the balanced equation, one can calculate the amount of reactants required to produce a desired amount of product, or vice versa. For instance, in the combustion of butane, stoichiometry helps us understand how many moles of oxygen would react with a given amount of butane to produce carbon dioxide and water.
Mole Calculation
Mole calculation is integral to stoichiometry as it connects the microscopic world of atoms and molecules to the macroscopic quantities that can be measured in the laboratory. A mole is defined as a unit that represents approximately 6.022 x 1023 particles, which is Avogadro's number. This large constant provides a bridge between the atomic scale and the real-world scale.
To perform mole calculations in a chemical equation, you need the molar masses of the substances involved, which are typically found on the periodic table. By using the molar mass as a conversion factor, you can switch between the mass of a substance and the amount in moles. This process is critical when determining ingredients for a reaction or analyzing product yields—and it was precisely used in calculating the moles of oxygen needed to react with butane in our exercise.
Combustion Reaction

Components of Combustion

A combustion reaction is a chemical process that occurs when a substance (typically a hydrocarbon) reacts with oxygen to produce energy in the form of heat and light. The general formula for a hydrocarbon combustion reaction is: Hydrocarbon + Oxygen → Carbon Dioxide + Water. These reactions are exothermic, meaning they release energy.

Characteristics and Applications

Combustion of fuel, like butane in this exercise, is a practical application of chemistry in everyday life, powering engines, heaters, and stoves. Correctly balancing a combustion reaction is important for predicting the amounts of products formed and for ensuring a reaction goes to completion with no excess reactants.
Chemical Reaction Quantities
Understanding chemical reaction quantities is essential when it comes to predicting how much of each substance is involved in a chemical reaction. When you balance a chemical equation, you are effectively setting the stage to calculate these quantities. The coefficients in the balanced equation give you the mole ratio of reactants to products.
For example, if you need to find out how much carbon dioxide will be produced after burning a known amount of butane, you look at the balanced chemical equation. In the combustion reaction from the exercise, a 2:13 ratio of butane to oxygen is established. This means that for every 2 moles of butane, 13 moles of oxygen are required. It is important to note that the quantities generated or consumed in a reaction are constrained by the limiting reactant, which in this case is the butane, as you are calculating how much oxygen is needed to react with a specific amount of butane.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction shown, calculate how many moles of \(\mathrm{NO}_{2}\) form when each amount of reactant completely reacts. $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ (a) \(1.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (b) \(5.8 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (c) \(4.45 \times 10^{3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\) (d) \(1.006 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\)

The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) and calculate how many kilograms of carbon dioxide are added to the atmosphere per \(1.0 \mathrm{~kg}\) of octane burned. (Hint: Begin by writing a balanced equation for the combustion reaction.)

Hydrochloric acid can dissolve solid iron according to the reaction: $$ \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ What minimum mass of \(\mathrm{HCl}\) in grams dissolves a \(2.8\)-g iron bar on a padlock? How much \(\mathrm{H}_{2}\) is produced by the complete reaction of the iron bar?

Consider the reaction between sulfur trioxide and water: $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ A chemist allows \(61.5 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(11.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) to react. When the reaction is finished, the chemist collects \(54.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Lead ions can be precipitated from solution with \(\mathrm{NaCl}\) according to the reaction: $$ \mathrm{Pb}^{2+}(a q)+2 \mathrm{NaCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{Na}^{+}(a q) $$ When \(135.8 \mathrm{~g}\) of \(\mathrm{NaCl}\) are added to a solution containing \(195.7 \mathrm{~g}\) of \(\mathrm{Pb}^{2+}, \mathrm{a} \mathrm{PbCl}_{2}\) precipitate forms. The precipitate is filtered and dried and found to have a mass of \(252.4 \mathrm{~g}\). Determine the limiting reactant, theoretical yield of \(\mathrm{PbCl}_{2}\), and percent yield for the reaction.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free