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How many of the following are found in 15.0 kmol of xylene \(\left(\mathrm{C}_{8} \mathrm{H}_{10}\right) ?\) (a) \(\mathrm{kg} \mathrm{C}_{8} \mathrm{H}_{10}\); (b) mol \(\mathrm{C}_{8} \mathrm{H}_{10}\); (c) lb-mole \(\mathrm{C}_{8} \mathrm{H}_{10} ;\) (d) mol (g-atom) \(\mathrm{C} ;\) (e) mol \(\mathrm{H} ;\) (f) \(\mathrm{g} \mathrm{C} ;\) (g) \(\mathrm{g} \mathrm{H} ;\) (h) molecules of \(\mathrm{C}_{8} \mathrm{H}_{10}\).

Short Answer

Expert verified
(a) 1592.4 kg of xylene, (b) 15000 mol of xylene, (c) 33.07 lb-mole of xylene, (d) 120000 mol (g-atom) of Carbon, (e) 150000 mol of Hydrogen, (f) 1441200 g of Carbon, (g) 151200 g of Hydrogen, (h) 9.033 x 10^28 molecules of xylene.

Step by step solution

01

Calculate Molecular Weight of Xylene.

First, we need to calculate the molecular weight of xylene (C8H10). For this, we use the atomic weights of Carbon (C) and Hydrogen (H). Carbon has an atomic weight of about 12.01 g/mol, and Hydrogen has an atomic weight of about 1.008 g/mol. So the total molecular weight of xylene is \(8 \times 12.01 + 10 \times 1.008 = 106.16 \, g/mol\).
02

Convert kmol to mol.

Now, we need to convert the given amount of xylene from kmol to mol. We know that 1 kmol equals 1000 mol. So in this problem, 15.0 kmol of xylene is \(15.0 \times 1000 = 15000 \, mol\).
03

Calculate mass in kg.

To calculate the mass in kg, we use the molecular weight of xylene and the amount in moles. The formula is \(Mass = moles \times molecular \, weight\). Thus, the mass of xylene would be \(15000 \, mol \times 106.16 \, g/mol \times 1 kg/1000 g = 1592.4 \, kg\).
04

Calculate amount in lb-mole.

To convert moles to lb-mole, we divide by a factor of 453.592 (since 1 lb-mole is equivalent to 453.592 moles). Thus, the amount of xylene is \(15000 \, mol / 453.592 = 33.07 \, lb-mole\).
05

Calculate amount in mol (g-atom) Carbon.

Xylene has 8 Carbon atoms per molecule. Therefore, the amount in mol (g-atom) of Carbon is \(15000 \, mol \times 8 = 120000 \, mol\).
06

Calculate amount in mol Hydrogen.

Similar to step 5, xylene has 10 Hydrogen atoms per molecule. Therefore, the amount in mol of Hydrogen is \(15000 \, mol \times 10 = 150000 \, mol\).
07

Calculate mass in g of Carbon and Hydrogen.

The mass in grams of Carbon can be calculated by \(120000 \, mol \times 12.01 \, g/mol = 1441200 \, g\). And the mass in grams of Hydrogen can be calculated by \(150000 \, mol \times 1.008 \, g/mol = 151200 \, g\).
08

Calculate number of molecules.

Using Avogadro's number (6.022 x 10^23 molecules/mol), the amount of xylene molecules is \(15000 \, mol \times 6.022 \times 10^{23} molecules/mol = 9.033 \times 10^{28} molecules\).

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Key Concepts

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

Molecular Weight Calculation
Understanding molecular weight calculation is essential for delving into stoichiometry problems. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. It's calculated by multiplying the atomic weight of each element by the number of atoms of that element in the molecule and then adding these values together.

For example, xylene has a chemical formula of \( C_8H_{10} \). The atomic weight of carbon (C) is approximately 12.01 g/mol, and hydrogen (H) is about 1.008 g/mol. The molecular weight of xylene is computed as follows: \((8 \times 12.01) + (10 \times 1.008) = 106.16 \, g/mol\).

Knowing the precise molecular weight is critical because it relates the physical mass of a substance to its amount in moles, a standard way of expressing quantities in chemistry. This is helpful when you're scaling reactions up or down or when converting between mass and moles, which is common in stoichiometry calculations.
Mole Concept
The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we experience every day. A mole is simply a counting unit similar to a dozen, except instead of 12, it represents approximately 6.022 \(\times\) 10^23 items, which is Avogadro's number.

This huge number corresponds to the number of carbon atoms in exactly 12 grams of pure carbon-12, and it's used because atoms and molecules are exceedingly tiny and numerous. When we say we have a mole of a substance, we mean we have Avogadro's number of molecules or atoms of that substance. The mole concept allows chemists to work with the submicroscopic particles in more manageable quantities of grams and kilograms, which can be weighed and handled in the laboratory.
Mass-to-Mole Conversion
Mass-to-mole conversion is at the heart of stoichiometry in chemistry. This conversion uses the molecular weight of a substance and the mole concept to interconvert between the mass of a substance and the number of moles.

The formula used is: \[\text{Number of moles} = \frac{\text{Mass (g)}}{\text{Molecular weight (g/mol)}}\]. For instance, if you have 1592.4 grams of xylene, and the molecular weight of xylene is 106.16 g/mol, the number of moles is calculated as \(\frac{1592.4}{106.16} = 15\) moles. This step is indispensable because reactions in chemistry are typically written on a mole basis, meaning you need to know the moles to understand how much of each reactant is needed and how much of each product will be made.

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Most popular questions from this chapter

The little-known rare earth element nauseum (atomic weight \(=172\) ) has the interesting property of being completely insoluble in everything but 25 -year- old single-malt Scotch. This curious fact was discovered in the laboratory of Professor Ludwig von Schlimazel, the eminent German chemist whose invention of the bathtub ring won him the Nobel Prize. Having unsuccessfully tried to dissolve nauseum in 7642 different solvents over a 10 -year period, Schlimazel finally came to the \(30 \mathrm{mL}\) of The Macsporran that was the only remaining liquid in his laboratory. Always willing to suffer personal loss in the name of science, Schlimazel calculated the amount of nauseum needed to make up a 0.03 molar solution, put the Macsporran bottle on the desk of his faithful technician Edgar P. Settera, weighed out the calculated amount of nauseum and put it next to the bottle, and then wrote the message that has become part of history: "Ed Settera. Add nauseum/" How many grams of nauseum did he weigh out? (Neglect the change in liquid volume resulting from the nauseum addition.)

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5\% and 15\%. A mixture containing 9.0 mole\% methane in air flowing at a rate of 7.00 \(\times 10^{2} \mathrm{kg} / \mathrm{h}\) is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (Note: Air may be taken to consist of \(\left.21 \text { mole } \% \mathrm{O}_{2} \text { and } 79 \% \mathrm{N}_{2} \text { and to have an average molecular weight of } 29.0 .\right)\)

In September 2014 the average price of gasoline in France was 1.54 euro/liter, and the exchange rate was \(\$ 1.29\) per euro ( \(\epsilon\) ). How much would you have paid, in dollars, for 50.0 kg of gasoline in France, assuming gasoline has a specific gravity of \(0.71 ?\) What would the same quantity of gasoline have cost in the United States at the prevailing average price of \(\$ 3.81 /\) gal?

Limestone (calcium carbonate) particles are stored in \(50-\mathrm{L}\) bags. The void fraction of the particulate matter is 0.30 (liter of void space per liter of total volume) and the specific gravity of solid calcium carbonate is 2.93. (a) Estimate the bulk density of the bag contents ( \(\mathbf{k g}\) CaCO \(_{3}\) /iter of total volume). (b) Estimate the weight ( \(W\) ) of the filled bags. State what you are neglecting in your estimate. (c) The contents of three bags are fed to a ball mill, a device something like a rotating clothes dryer containing steel balls. The tumbling action of the balls crushes the limestone particles and turns them into a powder. (See pp. \(21-64\) of Perry's Chemical Engineers' Handbook, 8 th ed.) The limestone coming out of the mill is put back into \(50-\mathrm{L}\) bags. Would the limestone (i) just fill three bags, (ii) fall short of filling three bags, or (iii) fill more than three bags? Briefly explain your answer.

As will be discussed in detail in Chapter \(5,\) the ideal-gas equation of state relates absolute pressure, \(P(\mathrm{atm}) ;\) gas volume, \(V(\text { liters }) ;\) number of moles of gas, \(n(\mathrm{mol}) ;\) and absolute temperature, \(T(\mathrm{K}):\) $$P V=0.08206 n T$$ (a) Convert the equation to one relating \(P(\mathrm{psig}), V\left(\mathrm{ft}^{3}\right), n(\mathrm{lb}-\mathrm{mole}),\) and \(T\left(^{\circ} \mathbf{F}\right)\). (b) \(\mathrm{A} 30.0\) mole \(\%\) CO and 70.0 mole \(\% \mathrm{N}_{2}\) gas mixture is stored in a cylinder with a volume of \(3.5 \mathrm{ft}^{3}\) at a temperature of \(85^{\circ} \mathrm{F}\). The reading on a Bourdon gauge attached to the cylinder is 500 psi. Calculate the total amount of gas (lb- mole) and the mass of \(\mathrm{CO}\left(\mathrm{Ib}_{\mathrm{m}}\right)\) in the tank. (c) Approximately to what temperature \(\left(^{\circ} \mathrm{F}\right)\) would the cylinder have to be heated to increase the gas pressure to 3000 psig, the rated safety limit of the cylinder? (The estimate would only be approximate because the ideal gas equation of state would not be accurate at pressures this high.)

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