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Chapter 4: Problem 143

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

Short Answer

Expert verified
The element X is Selenium (Se), and the compound H2X is called Hydroselenic Acid (H2Se). The mass of the H2X (H2Se) sample used is 0.2522 g.

Step by step solution

01

Identifying the element X by its total electrons

The ion X^(2-) has a total of 36 electrons; it means the neutral atom of element X has 34 electrons in its structure as the ion has added two extra electrons, so we have to find an element with an atomic number of 34. The element with atomic number 34 is Selenium (Se).
02

Naming the compound H2X

Now that we know X is Selenium, we can name the compound H2X. It is an acid with two hydrogen atoms and one selenium atom, forming a compound H2Se. This compound is called Hydroselenic Acid.
03

Finding the moles of OH- required for neutralization

To figure out the mass of H2X, we first need to find the number of moles of OH- ions required for complete neutralization. We will use the formula: \[ moles (OH^-) = Molarity\ (OH^-) \times Volume\ (OH^-) \] Plugging in the given values, we get: \[ moles (OH^-) = 0.175 M \times 0.0356 L = 0.00623 mol \]
04

Balancing the chemical equation

Now, we have to balance the chemical equation for the reaction: \[ H_{2}X(aq) + OH^{-}(aq) \longrightarrow X^{2-}(aq) + H_{2}O(l) \] The balanced equation turns out to be: \[ H_{2}Se(aq) + 2OH^{-} (aq) \longrightarrow Se^{2-}(aq) + 2H_{2}O(l) \]
05

Calculating moles of H2X from moles of OH-

Using the balanced stoichiometry of the reaction, we can calculate the number of moles of H2X (H2Se) consumed in the process: \[ moles\ (H2Se) = \frac{moles\ (OH^-)}{2} = \frac{0.00623\ mol}{2} = 0.003115\ mol \]
06

Computing mass of H2X from moles of H2X

Finally, to find the mass of the H2Se sample used, we multiply the number of moles by its molar mass: \[ mass\ (H2Se) = moles\ (H2Se) \times Molar\ mass\ (H2Se) \] The molar mass of H2Se is obtained by adding the molar masses of 2 Hydrogen atoms and 1 Selenium atom: \[ Molar\ mass\ (H2Se) = 2 \times Molar\ mass\ (H) + Molar\ mass\ (Se) = 2 \times 1.01 g/mol + 78.97 g/mol \] \[ Molar\ mass\ (H2Se) = 2.02\ g/mol + 78.97\ g/mol = 80.99\ g/mol \] Now, we multiply the obtained moles of H2Se by the molar mass: \[ mass (H2Se) = 0.003115\ mol \times 80.99\ g/mol = 0.252\,2\ g \] The mass of the H2X (H2Se) sample used is 0.2522 g.

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Most popular questions from this chapter

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl} ?\)

You are given a \(1.50-\mathrm{g}\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{~mL}\) of water and then add an excess of \(0.500 \mathrm{M}\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of \(0.641 \mathrm{~g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

Hydrochloric acid \((75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M})\) is added to \(225.0 \mathrm{~mL}\) of \(0.0550 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

A mixture contains only sodium chloride and potassium chloride. A \(0.1586-\mathrm{g}\) sample of the mixture was dissolved in water. It took \(22.90 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?

Write the balanced formula equation for the acid-base reactions that occur when the following are mixed. a. potassium hydroxide (aqueous) and nitric acid b. barium hydroxide (aqueous) and hydrochloric acid c. perchloric acid \(\left[\mathrm{HClO}_{4}(a q)\right]\) and solid iron(III) hydroxide d. solid silver hydroxide and hydrobromic acid e. aqueous strontium hydroxide and hydroiodic acid
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