Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 119

Consider the combustion reaction between 25.0 \(\mathrm{mL}\) of liquid methanol (density \(=0.850 \mathrm{g} / \mathrm{mL} )\) and 12.5 \(\mathrm{L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Short Answer

Expert verified
The volume of liquid water formed by condensing the water vapor produced in the combustion reaction between 25.0 mL of liquid methanol and 12.5 L of oxygen gas at STP is 13.4 mL.

Step by step solution

01

Write the balanced chemical equation

The combustion reaction of methanol (CH3OH) involves combining it with oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O). First, write down the reactants and products: CH3OH + O2 → CO2 + H2O To balance the equation, we need to make sure the number of atoms for each element is the same on both sides: CH3OH + 3/2 O2 → CO2 + 2 H2O Now, the balanced chemical equation for the combustion of methanol is: 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
02

Find the moles of reactants

To perform stoichiometry, we need the number of moles of each reactant. Given the volume and density of liquid methanol, we can find its mass and then its moles. As for the oxygen gas, its volume is given at STP (standard temperature and pressure), so we can use the molar volume of a gas at STP (22.4 L/mol) to convert the volume to moles: Moles of methanol: \(M_{CH3OH} = 25.0\,mL * 0.850\,g/mL = 21.25\,g\) \\ Molar mass of CH3OH = 32.04 g/mol \\ Moles of CH3OH: \(n_{CH3OH} = \frac{21.25\,g}{32.04\,g/mol} = 0.663\,mol\) Moles of oxygen: Molar volume of a gas at STP = 22.4 L/mol \\ Moles of O2: \(n_{O2} = \frac{12.5\,L}{22.4\,L/mol} = 0.558\,mol\)
03

Determine the limiting reactant

Use stoichiometry to compare the moles of reactants and find which is limiting: Moles of O2 needed for complete reaction of CH3OH: \(0.663\,mol\,CH3OH * \frac{3\,mol\,O2}{2\,mol\,CH3OH} = 0.995\,mol\,O2\) Since we only have 0.558 mol of O2, oxygen is the limiting reactant.
04

Calculate the moles of water formed

Use stoichiometry to find the number of moles of water formed, based on the limiting reactant (O2): Moles of H2O: \(0.558\,mol\,O2 * \frac{4\,mol\,H2O}{3\,mol\,O2} = 0.744\,mol\,H2O\)
05

Convert moles of water vapor to volume of liquid water

To find the volume of liquid water, we can use the density of water (1.00 g/mL) and the molar mass of water (18.02 g/mol): Mass of H2O: \(0.744\,mol\,H2O * 18.02\,g/mol = 13.4\,g\,H2O\) \\ Volume of liquid H2O: \(V_{H2O} = \frac{13.4\,g}{1.00\,g/mL} = 13.4\,mL\) The volume of liquid water formed by condensing the water vapor produced in the reaction is 13.4 mL.

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.

Start your free trial

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

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

A plasma-screen TV contains thousands of tiny cells filled with a mixture of \(\mathrm{Xe}, \mathrm{Ne}\) , and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, \(0.900 \mathrm{mm} \times 0.300 \mathrm{mm} \times 10.0 \mathrm{mm},\) contains 4\(\%\) Xe in a 1: Ne: He mixture at a total pressure of 500 torr. Calculate the number of Xe, Ne, and He atoms in the cell and state the assumptions you need to make in your calculation.

An aerosol spray can with a volume of 250 \(\mathrm{mL}\) contains 2.30 \(\mathrm{g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23^{\circ} \mathrm{C}\) , what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can's label says that exposure to temperatures above \(130^{\circ}\) F may cause the can to burst. What is the pressure in the can at this temperature?

Which one or more of the following statements are true? \begin{equation}\begin{array}{l}{\text { (a) } \mathrm{O}_{2} \text { will effuse faster than } \mathrm{Cl}_{2} \text { . }} \\ {\text { (b) Effusion and diffusion are different names for the same }} \\ {\text { process. }} \\\ {\text { (c) Perfume molecules travel to your nose by the process of }} \\\ {\text { effusion. }} \\ {\text { (d) The higher the density of a gas, the shorter the mean }} \\ {\text { free path. }}\end{array}\end{equation}

An ideal gas at a pressure of 1.50 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.800 \(\mathrm{L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is 695 torr, what is the volume of the bulb that was originally filled with gas?

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into \((\mathbf{a})\) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of Hg.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation
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