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Chapter 3: Problem 61

What amount (moles) is represented by each of these samples? a. \(150.0 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) b. \(10.0 \mathrm{mg} \mathrm{NO}_{2}\) c. \(1.5 \times 10^{16}\) molecules of \(\mathrm{BF}_{3}\)

Short Answer

Expert verified
a. \(0.939 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) b. \(2.17 \times 10^{-4} \mathrm{~mol} \mathrm{NO}_{2}\) c. \(2.49 \times 10^{-8} \mathrm{~mol} \mathrm{BF}_{3}\)

Step by step solution

01

a. Calculate the moles of Fe₂O₃.

To calculate the moles of Fe₂O₃, use the formula: moles = mass / molar mass First, find the molar mass of Fe₂O₃: Fe = 55.85 g/mol (for 2 Fe atoms = 2 x 55.85 g/mol) O = 16.00 g/mol (for 3 O atoms = 3 x 16.00 g/mol) Molar mass of Fe₂O₃ = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.7 g/mol Now, determine the moles of Fe₂O₃: moles = 150.0 g / 159.7 g/mol = 0.939 mol
02

b. Calculate the moles of NO₂.

To calculate the moles of NO₂, use the formula: moles = mass / molar mass First, convert the mass of NO₂ from mg to g: 10.0 mg = 0.010 g Now, find the molar mass of NO₂: N = 14.01 g/mol O = 16.00 g/mol (for 2 O atoms = 2 x 16.00 g/mol) Molar mass of NO₂ = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol Determine the moles of NO₂: moles = 0.010 g / 46.01 g/mol ≈ 2.17 x 10^{-4} mol
03

c. Calculate the moles of BF₃.

To calculate the moles of BF₃, use Avogadro's number to convert the given number of molecules to moles: number of moles = number of molecules / Avogadro's number number of moles = (1.5 x 10^16 molecules) / (6.022 x 10^23 molecules/mol) ≈ 2.49 x 10^{-8} mol

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Most popular questions from this chapter

A potential fuel for rockets is a combination of \(\mathrm{B} \cdot \mathrm{H}_{9}\) and \(\mathrm{O}_{-}\) The two react according to the following balanced equation: $$ 2 \mathrm{~B}_{5} \mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{~B}_{2} \mathrm{O}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(g) $$ If one tank in a rocket holds \(126 \mathrm{~g} \mathrm{~B}_{5} \mathrm{H}_{9}\) and another tank holds \(192 \mathrm{~g} \mathrm{O}_{2}\), what mass of water can be produced when the entire contents of each tank react together?

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\), butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\), and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\). a. A sample of \(\mathrm{ABS}\) plastic contains \(8.80 \% \mathrm{~N}\) by mass. It took \(0.605 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) to react completely with a \(1.20-\mathrm{g}\) sample of \(\mathrm{ABS}\) plastic. Bromine reacts \(1: 1\) (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer? b. What are the relative numbers of each of the monomer units in this polymer?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

The most common form of nylon (nylon-6) is \(63.68 \%\) carbon. \(12.38 \%\) nitrogen, \(9.80 \%\) hydrogen, and \(14.14 \%\) oxygen. Calculate the empirical formula for nylon-6.

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.
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