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Chapter 3: Problem 69

A crude oil burned in electrical generating plants contains about \(1.2 \%\) sulfur by mass. When the oil burns, the sulfur forms sulfur dioxide gas: $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How many liters of \(\mathrm{SO}_{2}(d=2.60 \mathrm{~g} / \mathrm{L})\) are produced when \(1.00 \times 10^{4} \mathrm{~kg}\) of oil burns at the same temperature and pressure?

Short Answer

Expert verified
Answer: The volume of sulfur dioxide gas produced is approximately 9.23 x 10^4 L.

Step by step solution

01

Calculate the mass of sulfur in the crude oil

We are given that the percentage of sulfur in crude oil is \(1.2\%\) and the mass of crude oil burned is \(1.00\times10^4\,\text{kg}\). Multiply the mass of crude oil by the percentage of sulfur to find the mass of sulfur in the crude oil: $$\text{mass of sulfur} = (1.2\%) (1.00\times10^4\,\text{kg})$$.
02

Calculate the mass of sulfur dioxide gas produced

From the balanced chemical equation, we know that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide gas. Using the molar masses of sulfur (\(32.07\,\text{g/mol}\)) and sulfur dioxide (\(64.07\,\text{g/mol}\)), we can convert the mass of sulfur in the crude oil to the mass of sulfur dioxide produced: $$\frac{\text{mass of sulfur}}{32.07\,\text{g/mol}} = \frac{\text{mass of sulfur dioxide}}{64.07\,\text{g/mol}}$$.
03

Calculate the volume of sulfur dioxide gas produced

We are given the density of sulfur dioxide gas (\(2.60\,\text{g/L}\)) and the mass of sulfur dioxide gas produced. We can calculate the volume of sulfur dioxide gas using its mass and density: $$\text{volume of }\mathrm{SO}_{2} = \frac{\text{mass of sulfur dioxide}}{2.60\,\text{g/L}}$$. Now, we can combine all the steps to find the volume of sulfur dioxide gas produced when \(1.00\times10^4\,\text{kg}\) of crude oil is burned. #Solution#
04

Calculate the mass of sulfur in the crude oil

We need to calculate the mass of sulfur in the crude oil: $$\text{mass of sulfur} = (1.2\%) (1.00\times10^4\,\text{kg}) = 120\,\text{kg} = 1.20\times10^5\,\text{g}$$.
05

Calculate the mass of sulfur dioxide gas produced

Now, we use the stoichiometry of the given chemical equation in terms of molar masses to convert the mass of sulfur to the mass of sulfur dioxide produced: $$\frac{\text{mass of sulfur}}{32.07\,\text{g/mol}} = \frac{\text{mass of sulfur dioxide}}{64.07\,\text{g/mol}}$$ $$\text{mass of sulfur dioxide} = \frac{1.20\times10^5\,\text{g} \times 64.07\,\text{g/mol}}{32.07\,\text{g/mol}} = 2.40\times10^5\,\text{g}$$.
06

Calculate the volume of sulfur dioxide gas produced

Finally, we can calculate the volume of sulfur dioxide gas using its density: $$\text{volume of }\mathrm{SO}_{2} = \frac{\text{mass of sulfur dioxide}}{2.60\,\text{g/L}} = \frac{2.40\times10^5\,\text{g}}{2.60\,\text{g/L}} = 9.23\times10^4\,\text{L}$$. So, the volume of sulfur dioxide gas produced when \(1.00\times10^4\,\text{kg}\) of crude oil is burned at the same temperature and pressure is approximately \(9.23\times10^4\,\text{L}\).

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Most popular questions from this chapter

Oxyacetylene torches used for welding reach temperatures near \(2000^{\circ} \mathrm{C}\). The reaction involved in the combustion of acetylene is $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) Starting with \(175 \mathrm{~g}\) of both acetylene and oxygen, what is the theoretical yield, in grams, of carbon dioxide? (b) If \(68.5 \mathrm{~L}(d=1.85 \mathrm{~g} / \mathrm{L})\) of carbon dioxide is produced, what is the percent yield at the same conditions of temperature and pressure? (c) How much of the reactant in excess is unused? (Assume 100\% yield.)

WEB DDT (dichlorodiphenyltrichloroethane) was the first chlorinated insecticide developed. It was used extensively in World War II to eradicate the mosquitoes that spread malaria. Its use was banned in the United States in 1978 because of environmental concerns. DDT is made up of carbon, hydrogen, and chlorine atoms. When a \(5.000-\mathrm{g}\) sample of DDT is burned in oxygen, \(8.692 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(1.142 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are obtained. A second five-gram sample yields \(2.571 \mathrm{~g}\) of \(\mathrm{HCl}\). What is the simplest formula for DDT?

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