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Chapter 5: Problem 71

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and a pressure of \(2.00 \mathrm{~atm}\). The products of the reaction are \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

Short Answer

Expert verified
In the reaction, methanol is the limiting reactant. Hence, when calculating the moles of water formed from the given moles of methanol and oxygen, a total of \(2.66\, \mathrm{mol \, H_2O}\) are produced when the reaction goes to completion.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between methanol and oxygen is: \[\mathrm{2CH_3OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(g)}\]
02

Calculate the moles of methanol

We can find the moles of methanol using the given volume and density as follows: \[\textrm{mass of methanol} = \textrm{volume} \times \textrm{density}\] \[\textrm{mass of methanol} = 50.0 \, \textrm{mL} \times 0.850 \, \frac{\textrm{g}}{\textrm{mL}} = 42.5 \, \textrm{g}\] Now, using the molar mass of methanol (32.04 g/mol), we can calculate the moles of methanol: \[\textrm{moles of methanol} =\frac{42.5 \, \mathrm{g}}{32.04\, \mathrm{g/mol}} \approx 1.33\, \mathrm{mol}\]
03

Calculate the moles of oxygen

Use the ideal gas law to find the moles of oxygen. The ideal gas law is: \[PV=nRT\] Where: \(P = 2.00 \, \mathrm{atm}\), \(V = 22.8 \, \mathrm{L}\), \(R = 0.0821\, \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}}\) and \(T = 27^\circ \mathrm{C} + 273.15 = 300.15\, \mathrm{K}\). Then, solve for n: \[\mathrm{n = \frac{PV}{RT}}\] \[\mathrm{n = \frac{(2.00 \, \mathrm{atm})(22.8 \, \mathrm{L})}{(0.0821\, \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}})(300.15\, \mathrm{K})} \approx 1.13 \, \mathrm{mol}}\]
04

Determine the limiting reactant

Use the balanced chemical equation and the moles of methanol and oxygen calculated before to determine the limiting reactant: \[\frac{\mathrm{1.33\, mol \, CH_3OH}}{2}\, \mathrm{mol \, CH_3OH/mol \, O_2} = 0.665\, \mathrm{mol \, O_2}\] Since 0.665 mol of oxygen is needed for all the methanol to react, and 1.13 mol of oxygen is available, the methanol is the limiting reactant.
05

Calculate the number of moles of water formed

Using the stoichiometry of the reaction, calculate the moles of water formed based on the moles of the limiting reactant (methanol): \[\mathrm{1.33\, mol \, CH_3OH} \times \frac{4\, \mathrm{mol \, H_2O}}{2 \, \mathrm{mol \, CH_3OH}} = 2.66\, \mathrm{mol \, H_2O}\] Therefore, 2.66 moles of water are formed if the reaction goes to completion.

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Key Concepts

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

Limiting Reactant Determination
Understanding the concept of a limiting reactant is crucial for students studying stoichiometry. It refers to the reactant that will be completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed.

In the exercise, the limiting reactant was identified after calculating the moles of both reactants, methanol and oxygen. The balanced chemical equation shows a 2:3 molar ratio between methanol and oxygen. By comparing the stoichiometric ratio with the actual moles available, we identified methanol as the limiting reactant.

Tips for Limiting Reactant Determination

  • Always start with a balanced chemical equation to determine the stoichiometric ratios.
  • Calculate the moles of each reactant using given quantities and molar masses or the ideal gas law for gases.
  • Divide the actual moles of each reactant by its stoichiometric coefficient to find which reactant has the smallest ratio. This is your limiting reactant.
  • Remember that the limiting reactant will dictate the amount of product formed.
Balanced Chemical Equations
A balanced chemical equation is an expression of a chemical reaction with the number of atoms for each element equal on both sides of the equation. This represents the law of conservation of mass. Balancing chemical equations allows us to predict the amounts of products and reactants involved and is a foundation of stoichiometry.

In our specific problem, the balanced chemical equation is \[2\mathrm{CH_3OH(l)} + 3\mathrm{O_2(g)} \rightarrow 2\mathrm{CO_2(g)} + 4\mathrm{H_2O(g)}\]. This equation indicates that two moles of methanol react with three moles of oxygen to produce two moles of carbon dioxide and four moles of water vapor.

Essential Points for Balanced Chemical Equations

  • Write down the unbalanced equation with all known reactants and products.
  • Use coefficients to balance the number of atoms of each element on both sides of the equation.
  • Check your work by counting atoms for consistency across the equation.
  • Understand that these coefficients represent the molar ratios used in stoichiometry calculations.
Ideal Gas Law Application
The ideal gas law is an equation of state for a hypothetical ideal gas. It is very useful for relating the pressure, volume, temperature, and amount of a gas with the formula \[PV = nRT\]. When we know any three of the four state variables, we can calculate the fourth.

In the context of the given exercise, we utilized the ideal gas law to determine the moles of oxygen. With known pressure, volume, and temperature, we rearranged the ideal gas law to solve for \(n\), the number of moles. This crucial step allowed us to proceed with identifying the limiting reactant.

Key Application Tips of Ideal Gas Law

  • Ensure temperature is in Kelvin for calculations.
  • Confirm that the units for pressure and volume match the units of the gas constant \(R\).
  • Solve for the unknown variable after confirming all other variables are known and properly converted.
  • Remember that real gases deviate from ideal behavior under high pressure or low temperature conditions.

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Most popular questions from this chapter

The average lung capacity of a human is \(6.0 \mathrm{~L}\). How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{~K}, P=1.00 \mathrm{~atm})\). b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{~K}, P=1.97 \mathrm{~atm})\). c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{~atm})\).

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3} .\) This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In 1894 , A. Combes (Comptes Rendus 1894, p. 1221\()\) reacted beryllium with the anion \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3}\). Show how Combes's data help to confirm that beryllium is a divalent metal.

An \(11.2-\mathrm{L}\) sample of gas is determined to contain \(0.50 \mathrm{~mol} \mathrm{~N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

A glass vessel contains \(28 \mathrm{~g}\) nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding enough mercury to fill one-half the container. b. Raising the temperature of the container from \(30 .{ }^{\circ} \mathrm{C}\) to \(60 .{ }^{\circ} \mathrm{C}\). c. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). d. Adding 28 g nitrogen gas.

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess \(\mathrm{HCl}\) gas to produce \(3.22 \mathrm{~L} \mathrm{H}_{2}(g)\) at \(373 \mathrm{~K}\) and \(0.951 \mathrm{~atm}\) and a manganese chloride compound \(\left(\mathrm{MnCl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?
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