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Chapter 4: Problem 43

Write a balanced molecular equation and a net ionic equation for the reaction that occurs when (a) solid \(\mathrm{CaCO}_{3}\) reacts with an aqueous solution of nitric acid; (b) solid iron(II) sulfide reacts with an aqueous solution of hydrobromic acid.

Short Answer

Expert verified
The balanced molecular equations and net ionic equations for the given reactions are: (a) CaCO3 and HNO3 reaction: Balanced molecular equation: \(CaCO_{3}(s) + 2HNO_{3}(aq) \rightarrow Ca(NO_{3})_{2}(aq) + H_{2}O(l) + CO_{2}(g)\) Net ionic equation: \(CaCO_{3}(s) + 2H^{+}(aq) \rightarrow Ca^{2+}(aq) + H_{2}O(l) + CO_{2}(g)\) (b) FeS and HBr reaction: Balanced molecular equation: \(FeS(s) + 2HBr(aq) \rightarrow FeBr_{2}(aq) + H_{2}S(g)\) Net ionic equation: \(FeS(s) + 2H^{+}(aq) \rightarrow Fe^{2+}(aq) + H_{2}S(g)\)

Step by step solution

01

Write the Unbalanced Equation

We need to start by writing an unbalanced chemical equation of the reaction. This is the equation that describes the reaction between solid \(\mathrm{CaCO}_{3}\) and aqueous nitric acid (HNO3): \(\mathrm{CaCO}_{3}(s) + \mathrm{HNO}_{3}(aq) \rightarrow\) Next, we need to predict the products of this reaction. \(\mathrm{CaCO}_{3}\) is a carbonate, so it will react with an acid to produce a salt, water, and carbon dioxide. In this case, the salt will be calcium nitrate, which has the formula \(\mathrm{Ca(NO}_{3})_{2}\): \(\mathrm{CaCO}_{3}(s) + \mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Ca(NO}_{3})_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)\)
02

Balance the Equation

Now we will balance the equation by making sure that we have the same number of each element on both sides of the equation: \(\mathrm{CaCO}_{3}(s) + 2\mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Ca(NO}_{3})_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)\) #a. Net Ionic Equation for CaCO3 and HNO3 Reaction#
03

Split the Compounds into Ions

Now, we will split the soluble compounds into their respective ions. Solid \(\mathrm{CaCO}_{3}\) and liquid water do not dissociate, so they will remain the same: \(\mathrm{CaCO}_{3}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)\)
04

Write the Net Ionic Equation

In the net ionic equation, we will cancel out the spectator ions (those which do not participate in the reaction): \(\mathrm{CaCO}_{3}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)\) The net ionic equation is: \(\mathrm{CaCO}_{3}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)\) Now, let's write the balanced molecular equation and net ionic equation for the reaction between solid iron(II) sulfide and an aqueous solution of hydrobromic acid. #b. Balanced Molecular Equation for FeS and HBr Reaction#
05

Write the Unbalanced Equation

First, write the unbalanced chemical equation for the reaction between solid iron(II) sulfide (FeS) and aqueous hydrobromic acid (HBr): \(\mathrm{FeS}(s) + \mathrm{HBr}(aq) \rightarrow\) This reaction will produce a salt, iron(II) bromide (FeBr2), and hydrogen sulfide, a gas (H2S): \(\mathrm{FeS}(s) + \mathrm{HBr}(aq) \rightarrow \mathrm{FeBr}_{2}(aq) + \mathrm{H}_{2}\mathrm{S}(g)\)
06

Balance the Equation

Now, balance the equation: \(\mathrm{FeS}(s) + 2\mathrm{HBr}(aq) \rightarrow \mathrm{FeBr}_{2}(aq) + \mathrm{H}_{2}\mathrm{S}(g)\) #b. Net Ionic Equation for FeS and HBr Reaction#
07

Split the Compounds into Ions

As before, we will split the soluble compounds into their respective ions: \(\mathrm{FeS}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{Br}^{-}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + 2\mathrm{Br}^{-}(aq) + \mathrm{H}_{2}\mathrm{S}(g)\)
08

Write the Net Ionic Equation

Now, we will cancel out the spectator ions (in this case, the bromide ions): \(\mathrm{FeS}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{S}(g)\) The net ionic equation for this reaction is: \(\mathrm{FeS}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{S}(g)\)

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Most popular questions from this chapter

Suppose you have a solution that might contain any or. all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+}\), and \(\mathrm{Mn}^{2+}\). Addition of \(\mathrm{HCl}\) solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resultant solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}_{3}\), (b) \(1.3 \times 10^{-2} \mathrm{M} \mathrm{MgSO}_{4},(\mathrm{c}) 0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (d) a mixture of \(45.0 \mathrm{~mL}\) of \(0.272 \mathrm{M} \mathrm{NaCl}\) and \(65.0 \mathrm{~mL}\) of \(0.0247 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\). Assume that the volumes are additive.

Which of the following solutions has the largest concentration of solvated protons: (a) \(0.2 \mathrm{MLiOH}\), (b) \(0.2 \mathrm{M} \mathrm{HI}\), (c) \(1.0 \mathrm{M}\) methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) ? Explain.

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) \(\mathrm{SrSO}_{4}(\mathrm{~s})+2 \mathrm{HNO}_{3}(a q)\) \(\mathrm{Zn}(s)+10 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow\) \(4 \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2} \mathrm{O}(g)+5 \mathrm{H}_{2} \mathrm{O}(l)\)

The safe drinking water standard for arsenic (which is usually found as arsenate, see \(4.115)\) is 50 parts per billion (ppb) in most developing countries. (a) How many grams of sodium arsenate are in 55 gallons of water, if the concentration of arsenate is 50 ppb? (b) In 1993 , naturally occurring arsenic was discovered as a major contaminant in the drinking water across the country of Bangladesh. Approximately 12 million people in Bangladesh still drink water from wells that have higher concentrations of arsenic than the standard. Recently, a chemistry professor from George Mason University was awarded a \(\$ 1\) million Grainger Challenge Prize for Sustainability for his development of a simple, inexpensive system for filtering naturally occuring arsenic from drinking water. The system uses buckets of sand, cast iron, activated carbon, and wood chips for trapping arsenic-containing minerals. Assuming the efficiency of such a bucket system is \(90 \%\) (meaning, \(90 \%\) of the arsenic that comes in is retained in the bucket and \(10 \%\) passes out of the bucket), how many times should water that is 500 ppb in arsenic be passed through to meet the 50 ppb standard?
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