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Chapter 7: Problem 107

Reducing \(\mathrm{SO}_{2}\) Emissions One way in which \(\mathrm{SO}_{2}\) is removed from the "stack" gases of coal-burning power plants is by spraying the gases with fine particles of solid calcium oxide suspended in \(\mathrm{O}_{2}\) gas. The product of the reaction of \(\mathrm{SO}_{2}\) \(\mathrm{CaO},\) and \(\mathrm{O}_{2}\) is calcium sulfate. a. Write a balanced chemical equation for this reaction. b. How many metric tons of calcium sulfate would be produced from each ton of \(\mathrm{SO}_{2}\) that is trapped?

Short Answer

Expert verified
Answer: 2125 kg of calcium sulfate is produced from trapping 1 metric ton of sulfur dioxide.

Step by step solution

01

Write the balanced chemical equation

Identify the reactants and products in this reaction between \(\mathrm{SO}_{2}\), \(\mathrm{CaO}\), and \(\mathrm{O}_{2}\) to produce calcium sulfate. This can be written as: $$\mathrm{SO}_{2} + \mathrm{CaO} + \frac{1}{2}\mathrm{O}_{2} \longrightarrow \mathrm{CaSO}_{4}$$
02

Calculate the molar mass of the reactants and products

In order to determine the amount of calcium sulfate produced from each ton of \(\mathrm{SO}_{2}\), we will need to use their molar masses. Molar mass of $\mathrm{SO}_{2} = 32 + 16(2) = 64\,\mathrm{g/mol}; Molar mass of $\mathrm{CaO} = 40 + 16 = 56\,\mathrm{g/mol}; Molar mass of \(\mathrm{CaSO}_{4} = 40 + 32 + 16(4) = 136\,\mathrm{g/mol}\).
03

Determining the mole ratio of \(\mathrm{SO}_{2}\) and \(\mathrm{CaSO}_{4}\)

From the balanced equation, the mole ratio of \(\mathrm{SO}_{2}\): \(\mathrm{CaSO}_{4}\) is 1:1. This means that one mole of \(\mathrm{SO}_{2}\) reacts to produce one mole of \(\mathrm{CaSO}_{4}\).
04

Calculate the amount of \(\mathrm{CaSO}_{4}\) produced from 1 metric ton of \(\mathrm{SO}_{2}\)

Now we have to find out how much \(\mathrm{CaSO}_{4}\) is produced when 1 metric ton (1000 kg) of \(\mathrm{SO}_{2}\) is trapped. Since 1 mole of \(\mathrm{SO}_{2}\) produces 1 mole of \(\mathrm{CaSO}_{4}\), we need to convert the mass of \(\mathrm{SO}_{2}\) to moles and then to the mass of \(\mathrm{CaSO}_{4}\). Number of moles of \(\mathrm{SO}_{2}\) in 1 metric ton = $$\frac{1000\,\mathrm{kg}}{0.064\,\mathrm{kg/mol}} = \frac{1000000\,\mathrm{g}}{64\,\mathrm{g/mol}} = 15625\,\mathrm{mol}$$ As explained earlier, the proportion between \(\mathrm{SO}_{2}\) and \(\mathrm{CaSO}_{4}\) is 1:1, therefore, we have 15625 mol of \(\mathrm{CaSO}_{4}\) as well. Let's convert it to mass now: Mass of the \(\mathrm{CaSO}_{4}\) = 15625 mol * 136 g/mol = $$2125000\,\mathrm{g} = 2125\,\mathrm{kg}$$ Thus, for each metric ton (1000 kg) of \(\mathrm{SO}_{2}\) trapped, 2125 kg of calcium sulfate are produced.

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Key Concepts

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

Balanced Chemical Equations
A balanced chemical equation is crucial to understand the proportions in which reactants combine and products form. In this exercise, the reaction involves sulfur dioxide (SO\(_2\)) reacting with calcium oxide (CaO) and oxygen (O\(_2\)) to form calcium sulfate (CaSO\(_4\)). Writing the balanced chemical equation is the first step. Here,
  • Sulfur dioxide reacts with one molecule of calcium oxide and half a molecule of oxygen to produce calcium sulfate.
  • The balanced equation is: \( \mathrm{SO}_2 + \mathrm{CaO} + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CaSO}_4 \).
Balancing an equation ensures that the same number of each type of atom is on both sides, respecting the conservation of mass.
Molar Mass Calculation
Understanding molar mass is an important part of solving chemical reaction problems. Molar mass is the mass of one mole of a substance, and it allows us to convert between grams and moles. Here's how to calculate it for the substances involved:
  • Sulfur dioxide (SO\(_2\)): With sulfur's atomic mass at 32 and oxygen's at 16, the molar mass is \(32 + 16 \times 2 = 64\,\text{g/mol}\).
  • Calcium oxide (CaO): Calcium is 40 and oxygen is 16, giving us \(40 + 16 = 56\,\text{g/mol}\).
  • Calcium sulfate (CaSO\(_4\)): Add calcium (40), sulfur (32), and oxygen (16\times4) amounts to \(40 + 32 + 64 = 136\,\text{g/mol}\).
These calculations allow for determining how much reactant and product are needed or formed based on their masses.
Mole-to-Mole Conversions
Mole-to-mole conversions stem directly from the coefficients in a balanced chemical equation. They express the ratio in which reagents participate in a reaction. For this reaction:
  • The balanced equation showed that the mole ratio of SO\(_2\) to CaSO\(_4\) is 1:1.
This means that every mole of sulfur dioxide will produce one mole of calcium sulfate. This simple relationship helps simplify calculations when determining the amount of product formed from a given amount of reactant. Since both reactants and products are involved in a 1:1 ratio, the conversion is straightforward.
Calcium Sulfate Production
The process of calculating calcium sulfate production involves several steps.
  • First, the mass of trapped sulfur dioxide is converted to moles using its molar mass. For instance, 1 metric ton (1000 kg) equals 1,000,000 grams, which divided by 64 g/mol gives 15,625 moles of sulfur dioxide.
  • Given the 1:1 mole ratio, 15,625 moles of SO\(_2\) will produce 15,625 moles of CaSO\(_4\).
  • To find the mass of calcium sulfate, multiply the number of moles by its molar mass: 15,625 moles \(\times 136\,\text{g/mol} = 2,125,000\,\text{g}\) or 2,125 kg.
Thus, trapping 1 metric ton of sulfur dioxide results in the production of 2,125 kg of calcium sulfate.

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Most popular questions from this chapter

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