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Chapter 4: Problem 1

Balance the following equations by inspection. (a) \(\mathrm{SO}_{3} \longrightarrow \mathrm{SO}_{2}+\mathrm{O}_{2}\) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HClO}_{4}\) (c) \(\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HNO}_{3}+\mathrm{NO}\) (d) \(\mathrm{PCl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+\mathrm{HCl}\)

Short Answer

Expert verified
The balanced equations are: (a) \(2\mathrm{SO}_{3} \longrightarrow 2\mathrm{SO}_{2}+\mathrm{O}_{2}\) (b) \(2\mathrm{Cl}_{2} \mathrm{O}_{7} + 8\mathrm{H}_{2} \mathrm{O} \longrightarrow 8\mathrm{HClO}_{4}\) (c) \(3\mathrm{NO}_{2}+ \mathrm{H}_{2} \mathrm{O} \longrightarrow 2\mathrm{HNO}_{3}+ \mathrm{NO}\) (d) \( \mathrm{PCl}_{3} + 3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+ 3\mathrm{HCl}\)

Step by step solution

01

- Examining and Balancing equation (a)

In the given equation \(\mathrm{SO}_{3} \longrightarrow \mathrm{SO}_{2}+\mathrm{O}_{2}\), count the number of each element on both sides. On the left, there are 1 sulfur and 3 oxygen atoms. On the right side, there are 1 sulfur, 2 oxygen from \(\mathrm{SO}_{2}\) and 2 from \(\mathrm{O}_{2}\) totaling to 4 oxygen atoms. Place the coefficient 2 in front of \(\mathrm{SO}_{3}\) and \(\mathrm{SO}_{2}\) to balance the number of Sulfur and Oxygen atoms. The balanced chemical equation is \(2\mathrm{SO}_{3} \longrightarrow 2\mathrm{SO}_{2}+\mathrm{O}_{2}\).
02

- Examining and Balancing equation (b)

In the equation \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HClO}_{4}\), there are 2 Chlorine, 7 Oxygen and 2 Hydrogen atoms on the left side and on the right side, there is 1 Chlorine, 4 Oxygen and 1 Hydrogen atom. Adjust the coefficients to make the numbers of atoms the same on both sides of the equation. The balanced equation is \(2\mathrm{Cl}_{2} \mathrm{O}_{7} + 8\mathrm{H}_{2} \mathrm{O} \longrightarrow 8\mathrm{HClO}_{4}\).
03

- Examining and Balancing equation (c)

In the equation \(\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HNO}_{3}+\mathrm{NO}\), there are 1 Nitrogen, 2 Oxygen and 2 Hydrogen atoms on the left side and on the right side, there are 2 Nitrogen, 4 Oxygen and 1 Hydrogen atoms. Again, adjust coefficients so that the numbers of each type of atom match on both sides. The balanced equation is \(3\mathrm{NO}_{2}+ \mathrm{H}_{2} \mathrm{O} \longrightarrow 2\mathrm{HNO}_{3}+ \mathrm{NO}\).
04

- Examining and Balancing equation (d)

In the equation \(\mathrm{PCl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+\mathrm{HCl}\), there is 1 Phosphorus, 3 Chlorine and 2 Hydrogen atoms on the left side and on the right side, there is 1 Phosphorus, 1 Chlorine, 3 Hydrogen atoms from \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and 1 from \(\mathrm{HCl}\) totaling to 4 Hydrogen and 3 Oxygen atoms. Adjust the coefficients to balance the equation, which is \( \mathrm{PCl}_{3} + 3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+ 3\mathrm{HCl}\).

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Most popular questions from this chapter

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Given two liters of \(0.496 \mathrm{M} \mathrm{KCl},\) describe how you would use this solution to prepare \(250.0 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{KCl} .\) Give sufficient details so that another student could follow your instructions.

What volume of \(0.149 \mathrm{M} \mathrm{HCl}\) must be added to \(1.00 \times 10^{2} \mathrm{mL}\) of \(0.285 \mathrm{M} \mathrm{HCl}\) so that the resulting solution has a molarity of \(0.205 \mathrm{M} ?\) Assume that the volumes are additive.

A mixture contains only \(\mathrm{CuCl}_{2}\) and \(\mathrm{FeCl}_{3}\). A \(0.7391 \mathrm{g}\) sample of the mixture is completely dissolved in water and then treated with \(\mathrm{AgNO}_{3}(\) aq). The following reactions occur. $$\begin{aligned} &\mathrm{CuCl}_{2}(\mathrm{aq})+2 \mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow 2 \mathrm{AgCl}(\mathrm{s})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \end{aligned}$$ $$\begin{aligned} &\mathrm{FeCl}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow 3 \mathrm{AgCl}(\mathrm{s})+\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq})\end{aligned}$$ If it takes \(86.91 \mathrm{mL}\) of \(0.1463 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the chloride as \(\mathrm{AgCl}\), then what is the percentage by mass of copper in the mixture?
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