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Chapter 4: Problem 67

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

Short Answer

Expert verified
a. KOH + \(\mathrm{HNO}_{3}\) → H2O + KNO3 b. \(\mathrm{Ba(OH)}_{2}\) + 2HCl → 2H2O + BaCl2 c. 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) → 3H2O + Fe(ClO4)3 d. \(\mathrm{AgOH}\) + HBr → H2O + AgBr e. \(\mathrm{Sr(OH)}_{2}\) + 2HI → 2H2O + SrI2

Step by step solution

01

a. potassium hydroxide (aqueous) and nitric acid

1. Write the formulas of the reactants: potassium hydroxide (KOH) and nitric acid (\(\mathrm{HNO}_{3}\)). 2. Identify the products of the reaction: When an acid (H+) reacts with a base (OH-), they form water (H2O) and a salt. In this case, the salt formed is potassium nitrate (KNO3). 3. Write the unbalanced equation: KOH + \(\mathrm{HNO}_{3}\) → H2O + KNO3 4. Balance the equation: The equation is already balanced. The balanced formula equation for the reaction is: KOH + \(\mathrm{HNO}_{3}\) → H2O + KNO3
02

b. barium hydroxide (aqueous) and hydrochloric acid

1. Write the formulas of the reactants: barium hydroxide (\(\mathrm{Ba(OH)}_{2}\)) and hydrochloric acid (\(\mathrm{HCl}\)). 2. Identify the products of the reaction: Water (H2O) and the salt barium chloride (BaCl2). 3. Write the unbalanced equation: \(\mathrm{Ba(OH)}_{2}\) + HCl → H2O + BaCl2 4. Balance the equation: To balance the equation, we need 2 moles of hydrochloric acid to provide enough H+ ions for both OH- ions from barium hydroxide: \(\mathrm{Ba(OH)}_{2}\) + 2HCl → 2H2O + BaCl2 The balanced formula equation for the reaction is: \(\mathrm{Ba(OH)}_{2}\) + 2HCl → 2H2O + BaCl2
03

c. perchloric acid and solid iron(III) hydroxide

1. Write the formulas of the reactants: Perchloric acid (HClO4) and solid iron(III) hydroxide (\(\mathrm{Fe(OH)}_{3}\)). 2. Identify the products of the reaction: Water (H2O) and the salt iron(III) perchlorate (Fe(ClO4)3). 3. Write the unbalanced equation: HClO4 + \(\mathrm{Fe(OH)}_{3}\) → H2O + Fe(ClO4)3 4. Balance the equation: To balance the equation, we need 3 moles of perchloric acid: 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) → 3H2O + Fe(ClO4)3 The balanced formula equation for the reaction is: 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) → 3H2O + Fe(ClO4)3
04

d. solid silver hydroxide and hydrobromic acid

1. Write the formulas of the reactants: Solid silver hydroxide (\(\mathrm{AgOH}\)) and hydrobromic acid (\(\mathrm{HBr}\)). 2. Identify the products of the reaction: Water (H2O) and the salt silver bromide (AgBr). 3. Write the unbalanced equation: \(\mathrm{AgOH}\) + HBr → H2O + AgBr 4. Balance the equation: The equation is already balanced. The balanced formula equation for the reaction is: \(\mathrm{AgOH}\) + HBr → H2O + AgBr
05

e. aqueous strontium hydroxide and hydroiodic acid

1. Write the formulas of the reactants: Aqueous strontium hydroxide (\(\mathrm{Sr(OH)}_{2}\)) and hydroiodic acid (\(\mathrm{HI}\)). 2. Identify the products of the reaction: Water (H2O) and the salt strontium iodide (SrI2). 3. Write the unbalanced equation: \(\mathrm{Sr(OH)}_{2}\) + HI → H2O + SrI2 4. Balance the equation: To balance the equation, we need 2 moles of hydroiodic acid: \(\mathrm{Sr(OH)}_{2}\) + 2HI → 2H2O + SrI2 The balanced formula equation for the reaction is: \(\mathrm{Sr(OH)}_{2}\) + 2HI → 2H2O + SrI2

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Most popular questions from this chapter

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a \(0.1472-\mathrm{g}\) sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed \(0.2327 \mathrm{~g}\). Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the \(C R C\) Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

In most of its ionic compounds, cobalt is either \(\mathrm{Co}(\mathrm{II})\) or \(\mathrm{Co}(\mathrm{III})\). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of \(0.308 \mathrm{~g}\). A second sample of \(0.416 \mathrm{~g}\) of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was \(0.145 \mathrm{~g}\). a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

Specify which of the following are oxidation-reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)
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