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Chapter 20: Problem 49

An electric power station annually burns \(3.1 \times 10^{7} \mathrm{~kg}\) of coal containing 2.4 percent sulfur by mass. Calculate the volume of \(\mathrm{SO}_{2}\) emitted at STP.

Short Answer

Expert verified
The volume of \(\mathrm{SO}_{2}\) emitted at STP is \(5.19 \times 10^{8}\) L

Step by step solution

01

Determine the mass of sulfur

First, calculate the total mass of sulfur in coal by multiplying the mass of the coal with the percentage of sulfur. This gives us: \(3.1 \times 10^{7} \mathrm{~kg} \times 0.024 = 7.44 \times 10^{5}\) kilograms of sulfur
02

Calculate the moles of sulfur

Next, convert the mass determined above into moles using the molar mass (S = 32.065 g/mol). Remember to convert the mass of sulfur into grams first. We have: \((7.44 \times 10^{5} \mathrm{~kg} \times 1000) / 32.065\)= \((7.44 \times 10^{8}\) g) / 32.065\) g/mol = \(2.32 \times 10^{7}\) mol
03

Apply stoichiometry

According to the reaction equation \(S + O^{2}- > SO_{2}\), one mole of \(S\) produces one mole of \(SO_{2}\). Hence the moles of \(SO_{2}\) produced will be equal to the moles of \(S\), which is \(2.32 \times 10^{7}\) mol
04

Calculate the Volume of \(SO_{2}\)

At STP, one mole of gas occupies 22.4 liters. So the volume of \(SO_{2}\) can be found by multiplying the moles of \(SO_{2}\) by 22.4 L/mol: \(2.32 \times 10^{7} \mathrm{~mol} \times 22.4 \mathrm{~L/mol} = 5.19 \times 10^{8}\) L

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