Question

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

Answer: 922.46 liters

Step-by-step solution

Calculate the mass of sulfur in the crude oil

In order to determine the mass of sulfur in the crude oil, we use the percentage by mass: Mass of sulfur = (percentage mass of sulfur/100) * total mass of crude oil _mass_sulfur_ = (1.2/100) * (1.00 x 10^4 kg) Convert the mass of sulfur to grams: _mass_sulfur_ = (1.2/100) * (1.00 x 10^4 * 1000) g = 1200 g

Convert the mass of sulfur to moles

To find the number of moles of sulfur, we use the molar mass of sulfur (32.06 g/mol): Moles of sulfur = mass of sulfur/molar mass of sulfur _moles_sulfur_ = (1200 g)/(32.06 g/mol) = 37.42 mol

Determine the moles of SO2 produced using stoichiometry

According to the balanced equation, 1 mol of S reacts with 1 mol of O2 to produce 1 mol of SO2. So, the moles of SO2 produced will be equal to the moles of sulfur: _moles_SO2_ = _moles_sulfur_ = 37.42 mol

Calculate the mass of SO2 produced

Now we will find the mass of SO2 produced using its molar mass (64.06 g/mol). Mass of SO2 = moles of SO2 * molar mass of SO2 _mass_SO2_ = (37.42 mol) * (64.06 g/mol) = 2396.39 g

Calculate the volume of SO2 produced

Finally, we can find the volume of SO2 produced by dividing the mass of SO2 by its density (2.60 g/L): Volume of SO2 = mass of SO2 / density of SO2 _volume_SO2_ = (2396.39 g) / (2.60 g/L) = 922.46 L Thus, 922.46 liters of SO2 are produced when 1.00 x 10^4 kg of crude oil containing 1.2% sulfur is burned at the same temperature and pressure.