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Chapter 4: Problem 41

Write balanced molecular and net ionic equations for the following reactions, and identify the gas formed in each: (a) solid cadmium sulfide reacts with an aqueous solution of sulfuric acid; (b) solid magnesium carbonate reacts with an aqueous solution of perchloric acid.

Short Answer

Expert verified
(a) Balanced molecular equation: \(CdS (s) + H鈧係O鈧 (aq) \rightarrow CdSO鈧 (aq) + H鈧係 (g)\) Net ionic equation: \(CdS (s) + 2 H鈦 (aq) \rightarrow Cd虏鈦 (aq) + H鈧係 (g)\) Gas formed: Hydrogen sulfide (H鈧係) (b) Balanced molecular equation: \(MgCO鈧 (s) + 2 HClO鈧 (aq) \rightarrow Mg(ClO鈧)鈧 (aq) + CO鈧 (g)\) Net ionic equation: \(MgCO鈧 (s) + 2 H鈦 (aq) \rightarrow Mg虏鈦 (aq) + CO鈧 (g)\) Gas formed: Carbon dioxide (CO鈧)

Step by step solution

01

Write the unbalanced molecular equation

Cadmium sulfide (CdS) reacts with sulfuric acid (H鈧係O鈧) to form cadmium sulfate (CdSO鈧) and hydrogen sulfide (H鈧係) gas. Unbalanced molecular equation: CdS (s) + H鈧係O鈧 (aq) 鈫 CdSO鈧 (aq) + H鈧係 (g)
02

Balance the equation

The equation is already balanced as written: CdS (s) + H鈧係O鈧 (aq) 鈫 CdSO鈧 (aq) + H鈧係 (g)
03

Write the net ionic equation

First, split the aqueous compounds into their respective ions: CdS (s) + 2 H鈦 (aq) + SO鈧劼测伝 (aq) 鈫 Cd虏鈦 (aq) + SO鈧劼测伝 (aq) + H鈧係 (g) Next, remove the spectator ion (SO鈧劼测伝), and write the net ionic equation: CdS (s) + 2 H鈦 (aq) 鈫 Cd虏鈦 (aq) + H鈧係 (g) (b) Solid magnesium carbonate reacts with an aqueous solution of perchloric acid
04

Write the unbalanced molecular equation

Magnesium carbonate (MgCO鈧) reacts with perchloric acid (HClO鈧) to form magnesium perchlorate (Mg(ClO鈧)鈧) and carbon dioxide (CO鈧) gas. Unbalanced molecular equation: MgCO鈧 (s) + HClO鈧 (aq) 鈫 Mg(ClO鈧)鈧 (aq) + CO鈧 (g)
05

Balance the equation

To balance the equation, we need 2 moles of perchloric acid for every mole of magnesium carbonate: MgCO鈧 (s) + 2 HClO鈧 (aq) 鈫 Mg(ClO鈧)鈧 (aq) + CO鈧 (g)
06

Write the net ionic equation

First, split the aqueous compounds into their respective ions: MgCO鈧 (s) + 2 H鈦 (aq) + 2 ClO鈧勨伝 (aq) 鈫 Mg虏鈦 (aq) + 2 ClO鈧勨伝 (aq) + CO鈧 (g) Next, remove the spectator ion (ClO鈧勨伝), and write the net ionic equation: MgCO鈧 (s) + 2 H鈦 (aq) 鈫 Mg虏鈦 (aq) + CO鈧 (g)

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Most popular questions from this chapter

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\), and \(\mathrm{Fe}^{2+}\), which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+}\). If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of \(\mathrm{Na}^{+}\) are needed to replace these ions?

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the observations made above. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C}\), and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or acid-base reactions. (a) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(I)\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (d) \(\mathrm{SrSO}_{4}(\mathrm{~s})+2 \mathrm{HNO}_{3}(a q)\) \(\mathrm{Zn}(s)+10 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow\) \(4 \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2} \mathrm{O}(g)+5 \mathrm{H}_{2} \mathrm{O}(l)\)
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