/*! 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 36 Calculate the following quantiti... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the following quantities: (a) mass, in grams, of \(1.50 \times 10^{-2} \mathrm{~mol} \mathrm{CdS}\) (b) number of moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(86.6 \mathrm{~g}\) of this substance (c) number of molecules in \(8.447 \times 10^{-2} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{6}\) (d) number of \(\mathrm{O}\) atoms in \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Short Answer

Expert verified
(a) The mass of \(1.50 \times 10^{-2}\) mol CdS is 2.167 g. (b) The number of moles of NH4Cl in 86.6 g is 1.62 mol. (c) The number of molecules in \(8.447 \times 10^{-2}\) mol C6H6 is \(5.08 \times 10^{22}\) molecules. (d) The number of O atoms in \(6.25 \times 10^{-3}\) mol Al(NO3)3 is \(3.39 \times 10^{22}\) O atoms.

Step by step solution

01

(a) Calculate the mass of \(1.50 \times 10^{-2}\) mol CdS

To calculate the mass of \(1.50 \times 10^{-2}\) mol of CdS (cadmium sulfide), we'll first determine the molar mass of CdS. The molar mass of CdS = (molar mass of Cd) + (molar mass of S) = \(112.41 \mathrm{~g/mol}\) + \(32.07 \mathrm{~g/mol}\) = \(144.48 \mathrm{~g/mol}\) Now, we use the equation \(m = n \times M\) to calculate the mass of CdS. mass = \((1.50 \times 10^{-2} \mathrm{~mol~CdS}) \times (144.48 \mathrm{~g/mol}) = 2.167 \mathrm{~g~CdS}\)
02

(b) Calculate the number of moles of NH4Cl in 86.6 g of this substance.

Now, we will determine the number of moles of NH4Cl (ammonium chloride) in 86.6 g of the substance. Molar mass of NH4Cl = (molar mass of N) + 4 × (molar mass of H) + (molar mass of Cl) = \(14.01 \mathrm{~g/mol}\) + 4 × \(1.01 \mathrm{~g/mol}\) + \(35.45 \mathrm{~g/mol}\) = \(53.49 \mathrm{~g/mol}\) Using the formula \(n = m/M\), we can compute the number of moles of NH4Cl. number of moles = \(\frac{86.6 \mathrm{~g}}{53.49 \mathrm{~g/mol}} = 1.62 \mathrm{~mol~NH4Cl}\)
03

(c) Calculate the number of molecules in \(8.447 \times 10^{-2}\) mol C6H6.

To find the number of molecules in \(8.447 \times 10^{-2}\) mol of C6H6 (benzene), we use the formula \(n_\text{particles} = n \times N_A\). Number of molecules = \((8.447 \times 10^{-2} \mathrm{~mol~C6H6}) \times (6.022 \times 10^{23} \, \text{molecules/mol})\) = \(5.08 \times 10^{22} \mathrm{~molecules~C6H6}\)
04

(d) Calculate the number of O atoms in \(6.25 \times 10^{-3}\) mol Al(NO3)3.

First, we'll find the number of moles of O atoms in \(6.25 \times 10^{-3}\) mol of Al(NO3)3. For each Al(NO3)3 molecule, there are 3 times 3 O atoms (9 O atoms) in each NO3 group. number of moles of O atoms = \((6.25 \times 10^{-3} \mathrm{~mol~Al(NO3)3}) \times (9 \, \text{O atoms per Al(NO3)3 molecule})\) = \(5.625 \times 10^{-2} \mathrm{~mol~O}\) Now we use the formula \(n_\text{particles} = n \times N_A\) to find the number of O atoms. Number of O atoms = \((5.625 \times 10^{-2} \mathrm{~mol~O}) \times (6.022 \times 10^{23} \, \text{atoms/mol})\) = \(3.39 \times 10^{22} \mathrm{~O~atoms}\)

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Key Concepts

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

Moles and Molecules
The concept of moles and molecules is foundational in chemistry. A **mole** is a unit that measures quantity of particles, typically atoms or molecules, and it is represented by Avogadro's number, \(6.022 \times 10^{23}\). This means one mole of any substance contains exactly \(6.022 \times 10^{23}\) molecules or atoms. For practical purposes, the mole allows chemists to convert between atoms, molecules, and grams.

For example, if you have 0.08447 mol of benzene (\(C_6H_6\)), using Avogadro's number, we can find the total molecules in this sample as follows:
  • Given that the number of molecules \(n_{particles} = n \times N_A\)
  • Substitute, \(n_{particles} = 0.08447 \times 6.022 \times 10^{23}\)
This calculation will yield the total number of molecules, giving us a concrete idea of the molecular scale and abundance involved. Understanding moles and molecules streamlines measurements and computations in chemistry by bridging macroscopic and molecular scales.
Chemical Reactions
Chemical reactions are processes that convert reactants into products through the breaking and forming of bonds. Understanding chemical reactions requires knowledge of how quantities relate in reactions, usually expressed in chemical equations. Each component in an equation is expressed in moles, providing a way to convey the proportions of each substance involved.

For instance, to determine how many oxygen atoms are present in a compound like \(Al(NO_3)_3\), recognize that each molecule contains nine oxygen atoms. Once you know how much of \(Al(NO_3)_3\) you have, you can determine the total number of oxygen atoms by multiplying this quantity by the number of oxygen atoms in the formula.
  • Identify the number of moles (e.g., \(6.25 \times 10^{-3}\) mol of \(Al(NO_3)_3\))
  • Multiply by the number of O atoms per formula unit (9 in this case)
  • Compute total atoms using \(n \times N_A\)
Thus, chemical equations not only represent the reactants and products in a reaction but help to indicate the stoichiometric relationships necessary for complete conversions.
Stoichiometry
Stoichiometry is the field in chemistry that studies the quantitative relationships between substances in chemical reactions. It helps predict the amount of products that result from given reactants and is an essential tool in chemistry for practical and theoretical applications. It relies on mole ratios from balanced chemical equations.

Consider a reaction where we calculated the mass of \(CdS\) from moles. Begin by determining a compound's molar mass using atomic masses from the periodic table, then use this to convert moles to grams. For \(1.50 \times 10^{-2}\) moles of \(CdS\), calculate its mass using:
  • Find the molar mass of \(CdS\) (e.g., 144.48 g/mol)
  • Utilize the equation \(mass = n \times M\)
  • Substitute values: \(mass = (1.50 \times 10^{-2}) \times 144.48\)
This approach highlights the utility of stoichiometry in linking reactants to precise quantities of products, establishing it as a core competency for chemists in predicting and verifying chemical phenomena.

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Most popular questions from this chapter

Write balanced chemical equations for \((\mathbf{a})\) the complete combustion of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right),\) a common organic solvent; (b) the decomposition of solid mercury (I) carbonate into carbon dioxide gas, mercury, and solid mercury oxide; (c) the combination reaction between sulphur dioxide gas and liquid water to produce sulfurous acid.

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NO \(_{x}\) " gases. In 2009 , the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{x}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of \(500 \mathrm{~g}\) of octane.

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and HCl. (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is \(2.1 \mathrm{~g} / \mathrm{cm}^{3}\). It can be made by reacting boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3}\), with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2},\) and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

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