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Chapter 3: Problem 36

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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Most popular questions from this chapter

(a) What is the mass, in grams, of \(1.223 \mathrm{~mol}\) of iron(III) sulfate? (b) How many moles of ammonium ions are in \(6.955 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(1.50 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{\circ}\) ) if 0.05570 mol has a mass of \(15.86 \mathrm{~g}\) ?

The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}$.

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with $0.350 \mathrm{~kg}\( of \)\mathrm{Fe}_{2} \mathrm{O}_{3}$ (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.350 \mathrm{~kg}\) of $\mathrm{Fe}_{2} \mathrm{O}_{3}$ reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(42.1 \% \mathrm{Na}, 18.9 \% \mathrm{P}\), and \(39.0 \% \mathrm{O}\) (b) \(18.7 \% \mathrm{Li}, 16.3 \% \mathrm{C},\) and \(65.0 \% \mathrm{O}\) (c) \(60.0 \% \mathrm{C}, 4.4 \% \mathrm{H},\) and the remainder \(\mathrm{O}\)
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