Homework Solutions #2

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Homework is due in class by the date indicated above. Late homework will be docked significantly.

Q1

Complete the following chemical equations by writing down the correct coefficients.

\(Cl_{2\;(g)}+Fe_{(s)} \rightarrow FeCl_{3\;(s)}\)
\(Fe_2O_{3\;(s)}+H_{2\;(g)} \rightarrow Fe_{(s)}+H_2O_{(g)}\)
\(Al_{(s)}+ O_{2\;(g)} \rightarrow Al_2O_{3\;(s)}\)
\(KClO_{3\;(s)} \rightarrow KCl_{(s)}+ O_{2\;(q)}\)
\(Al(s) +CuSO_{4\;(aq)} \rightarrow Al_2(SO_4)_{3\;(aq)} +Cu_{(s)}\)

A3 100Q.

\(\underline{3}Cl_{2\;(g)}+\underline{2}Fe_{(s)} \rightarrow \underline{2}FeCl_{3\;(s)}\)
\(Fe_2O_{3\;(s)}+\underline{3}H_{2\;(g)} \rightarrow \underline{2}Fe_{(s)}+\underline{3}H_2O_{(g)}\)
\(\underline{4}Al_{(s)}+ \underline{3}O_{2\;(g)} \rightarrow \underline{2}Al_2O_{3\;(s)}\)
\(\underline{2}KClO_{3\;(s)} \rightarrow \underline{2}KCl_{(s)}+ \underline{3}O_{2\;(q)}\)
\(\underline{2}Al(s) + \underline{3}CuSO_{4\;(aq)} \rightarrow Al_2(SO_4)_{3\;(aq)} +\underline{3}Cu_{(s)}\)

H3 100Q.

Obey the “conservation of atoms numbers”.

Solution

When balancing a chemical equation, we should always obey the rule that all atoms present in the reactants must be accounted for among the products.

\(\underline{3}Cl_{2\;(g)}+\underline{2}Fe_{(s)} \rightarrow \underline{2}FeCl_{3\;(s)}\)

Confirm that 6 Cl atoms and 2 Fe atoms are on each side of the reaction

\(Fe_2O_{3\;(s)}+\underline{3}H_{2\;(g)} \rightarrow \underline{2}Fe_{(s)}+\underline{3}H_2O_{(g)}\)

Confirm that 2 Fe atoms 3 O atoms 6 H atoms are on each side of the reaction

\(\underline{4}Al_{(s)}+ \underline{3}O_{2\;(g)} \rightarrow \underline{2}Al_2O_{3\;(s)}\)

Confirm that 4 Al atoms 6 O atoms are on each side of the reaction

\(\underline{2}KClO_{3\;(s)} \rightarrow \underline{2}KCl_{(s)}+ \underline{3}O_{2\;(q)}\)

Confirm that 2 K atoms 1 Cl atom 6 O atoms are on each side of the reaction

\(\underline{2}Al(s) + \underline{3}CuSO_{4\;(aq)} \rightarrow Al_2(SO_4)_{3\;(aq)} +\underline{3}Cu_{(s)}\)

Confirm that 2 Al atoms 3 Cu atoms 1 S atom 12 O atoms are on each side of the reaction

Q2

Permanganate is the general name for a chemical compound containing the manganate(VII) ion, (\(MnO_4^−\)). Because manganese is in the +7 oxidation state, the permanganate(VII) ion is a strong oxidizing agent. For example, \(KMnO_4\) can will decompose upon heating to produce oxygen (via a disproportionation REOX reaction) via the following unbalanced chemical reaction:

\[KMnO_{4\;(s)} \rightarrow K_2MnO_{4\;(s)} + MnO_{2\;(s)} + O_{2\;(g)}\]

If 32.5g \(KMnO_4\) is reacted, what the mass of oxygen will be produced?

A3 106Q.

3.29 g

H3 106Q.

Balance the equation first and do stoichiometric calculation.

S3 106Q.

First, as with all stiochiometry problems, a balanced chemical equation is necessary:

\[\underline{2}KMnO_{4\;(s)} \rightarrow K_2MnO_{4\;(s)} + MnO_{2\;(s)} + O_{2\;(g)}\]

And we can see that for ever two moles of \(KMnO_4\) that are reacted, one mole of oxygen is formed.

Initial Moles of \(KMnO_4\)_:\[\dfrac{32.5 \;g}{(39.10 + 54.94 + 16.00 \times 4 \;g/mol)} = 0.206\; mol\]
Final Moles of oxygen (2:1 ratio to \(KMnO_4\)): \[\dfrac{1}{2} \times 0.206 \; \text{moles of} \; KMnO_4 =0.103\; \text{mol of}\; O_2\]
Mass of oxygen: mole of oxygen *32.00 g/mol = 3.29 g.

Q3

Consider the following reaction.

\[Mg(OH)_2 + 2 HCl \rightarrow MgCl_2 + 2 H_2O\]

Answer the following questions if 25.3 g of \(Mg(OH)_2\) were reacted with 22.5 g of \(HCl\):

What is the limiting reagent?
What is the theoretical yield of \(MgCl_2\)?
How many grams of excess are unreacted?
If the actual reaction produced 26.6 g of \(MgCl_2\), calculate the percent yield of generating \(MgCl_2\).

A3. 124A

\(HCl\)
29.4 g \(MgCl_2\)
7.3 g \(Mg(OH)_2\)
90.5%

H3. 124A

Percent Yield

S3. 124A

Molecular Mass of \(Mg(OH)_2\): M = 24.305 + (2*15.9994) + (2*1.00794) = 58.31968 g/mol
Molecular Mass of \(HCl\): M = 1.00794 + 35.453 = 36.46094 g/mol
Molecular Mass of \(MgCl_2\): M = 24.305 + (2*35.453) = 95.211 g/mol

\[Mg(OH)_2 + 2 HCl \rightarrow MgCl_2 + 2 H_2O\]

\[(25.3 \;g \;Mg(OH)_2) \times \dfrac{1\; mol\; Mg(OH)_2 }{ 58.31968\; g \;Mg(OH)_2 } = 0.434 \; mol \; Mg(OH)_2\]

\[ (22.5 \;g \;HCl) \times \dfrac{ 1\; mol\; HCl}{36.46094\; g \;HCl} = 0.617\; mol\; HCl\]

HAVE= 0.617 mol HCl / 0.434 mol Mg(OH)₂ = 1.42 mol HCl / 1 mol Mg(OH)₂
NEED= 2 mol HCl / 1 mol Mg(OH)₂

We have less \(HCl\) than what is needed. Therefore, \(HCl\) is the limiting reagent.

\[(0.617\; mol\; HCl) \times \dfrac{1\; mol\; MgCl_2 }{2\; mol\; HCl} \times \dfrac{95.211 \;g \;MgCl_2}{ 1\; mol\; MgCl_2} = 29.4\; g \;MgCl_2\]
\[(0.617 mol HCl) \times \dfrac{1\; mol\; Mg(OH)_2}{ 2\; mol\; HCl} \times \dfrac{58.31968\; g\; Mg(OH)_2}{ 1\; mol\; Mg(OH)_2} = 18.0 g \;Mg(OH)_2 \; \;reacted\].

\(Mg(OH)_2\) unreacted = 25.3 g – 18.0 g = 7.3 g

\[\text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \dfrac{26.6\; g \;MgCl_2 }{29.4\; g \;MgCl_2} \times 100 = 90.5\%\]

Q4

Write a balanced molecular equation for the dissociation of \(NaCl\) in water.
Based on the equation in part a, is \(NaCl\) a strong electrolyte or a weak electrolyte?
Hydrofluoric acid is a weak acid, and does not completely dissociate in water. Is it a strong electrolyte or a weak electrolyte?

H4.15S

How do strong electrolytes dissociate in water vs. weak electrolytes. http://chemwiki.ucdavis.edu/Under_Co...d_Electrolytes

S4.15S

NaCl dissociates completely, therefore \(NaCl\) breaks up into one positively charged sodium ion and one negatively charged chlorine ion.
When a compound in water dissociates completely into ions it is a strong electrolyte. This is because the resulting solution is able to more easily conduct electricity.
Hydrofluoric acid does not dissociate completely (weak acid), therefore a solution of \(HF\) and \(H_2O\) would not conduct electricity easily. Therefore it is a weak electrolyte.
A4.15S
\(NaCl \rightarrow Na^+ + Cl^-\)
\(NaCl\) is a strong electrolyte
Hydrofluoric acid is a weak electrolyte.

Q5

When both of the beakers are added together, chemical reaction takes place. Identify the balanced chemical reaction and the net ionic reaction.

A4.49D

\[AgNO_{3(aq)} + NaI_{(aq)} \rightarrow AgI_{(s)} + NaNO_{3 (aq)}\]

Ag⁺_(aq) + I^-_(aq) --> AgI_(s)

Na⁺ (the purple spheres) and NO₃^- (the pink spheres) are the spectator ions

NiBr₂_(aq) + 2KOH _(aq) --> Ni(OH)_{2 (s)} + 2KBr_(aq)

Ni²⁺ _(aq) + 2OH^- _(aq) --> Ni(OH)_{2 (s)}

Br^- (blue spheres) and K⁺ (orange spheres) are the spectator ions

MnSO₄_(aq)+ Na₂S _(aq) --> MnS_(s) + Na₂SO₄_(aq)

Mn²⁺ _(aq) + S²^-_(aq) --> MnS_(s)

SO₄ ^2- (cream spheres) and Na⁺ (blue spheres) are the spectator ions

Q6

What are the whole formula equations and the net ionic equations for the following reactions? Write No Reaction if no reactions occur

sodium chromate and zinc chloride
ammonium nitrate and potassium hydroxide
cesium sulfate and mercury(I) nitrate
lithium hydroxide and chromium(III) chloride
rubidium iodide and sodium bromide

A4.51J

\(CrO_{4\;(aq)}^{2-} + Zn_{(aq)}^{2+} \rightarrow ZnCrO_{4\;(s)}\)
No reaction
No reaction
\(3OH_{(aq)}^- + Cr_{(aq)}^{3+} \rightarrow Cr(OH)_{3\;(s)}\)
No reaction

H4.51J

Start by writing the overall chemical equation.
4.4_Ionic_Equations

S4.51J
a. Overall chemical equation:

\[Na_2 CrO_{4\;(aq)}+ZnCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)}+ZnCrO_{4\;(s)}\]

Complete ionic Equation:

\[2Na_{(aq)}^++ CrO_{4\;(aq)}^{2-} + Zn_{(aq)}^{2+} + 2Cl_{(aq)}^- \rightarrow 2Na_{(aq)}^+ + 2Cl_{(aq)}^- + ZnCrO_{4\; (s)}\]

Net Ionic Equation:

\[CrO_{4(aq)}^{2-}+Zn_{(aq)}^{2+} \rightarrow ZnCrO_{4\;(s)}\]

b. Overall chemical equation:

\[NH_4NO_{3\;(aq)} +KOH_{(aq)} \rightarrow NH_4OH_{(aq)} + KNO_{3\;(aq)} \]

Complete ionic equation:

\[NH_{4\;(aq)}^+ + NO_{3\;(aq)}^- + K_{(aq)}^+ + OH_{(aq)}^- \rightarrow NH_{4\;(aq)}^+ + OH_{(aq)}^- + K_{(aq)}^+ + NO_{3\;(aq)}^-\]

Net ionic equation:

At first it looks like all species cancel and no precipitation reaction occurs, but there is a coupledacid/base neutralization reaction between \(NH_4\) and \(OH^-\) to generate ammonia and water:

\[NH^+_{4\;(aq)} + OH^-_{(aq)} \rightleftharpoons H_2O_{(l)} + NH_3\]

c. Overall chemical equation:

\[Cs_2SO_{4(aq)}^2 + Hg_2(NO_3)_{2\;(aq)} \rightarrow 2CsNO_{3\;(aq)} +Hg_2 SO_{4(s)} \]

Complete Ionic Equation:

\[2Cs_{(aq)}^+ + SO_{4(aq)}^{2-}+Hg_{2(aq)}^{2+} + 2NO_{3(aq)}^- \rightarrow 2Cs_{(aq)}^++2NO_{3(aq)}^-+ 2Hg^+ + SO_{4\;(aq)}\]

Net Ionic Equation:

No reaction

d. Overall chemical Equation:

\[3LiOH_{(aq)}+CrCl_{3(aq)} \rightarrow 3LiCl_{(aq)}+Cr(OH)_{3(s)}\]

Complete ionic equation:

\[3Li_{(aq)}^++3OH_{(aq)}^-+Cr_{(aq)}^{3+}+3Cl_{(aq)}^- \rightarrow 3Li_{(aq)}^+ + 3Cl_{(aq)}^- + Cr(OH)_{3(s)}\]

Net ionic equation:

\[3OH_{(aq)}^-+Cr_{(aq)}^{3+} \rightarrow Cr(OH)_{3(s)}\]

e. Overall chemical equation:

\[RbI_{(aq)} +NaBr_{(aq)} \rightarrow RbBr_{(aq)} +NaI_{(aq)}\]

Complete Ionic equation:

\[Rb_{(aq)}^+ + I_{(aq)}^-+ Na_{(aq)}^+ + Br_{(aq)}^- \rightarrow Rb_{(aq)}^+ + Br_{(aq)}^- + Na_{(aq)}^+ + I_{(aq)}^-\]

Net Ionic Equation:

No reaction

Q7

If 55.6 mL of a 2.0 M solution is diluted to 181.6 mL, what is the final molarity of the solution?

A4.37B

0.61M

H4.37B

You will need to know how many moles are present in the solution. Then the new volume of solution will be needed.
http://chemwiki.ucdavis.edu/Wikitexts/ChemTutor/MOLS%2C_PERCENTS%2C_and_STOICHIOMETRY

S4.37B

To start, multiply the molarity with the volume of the original solution. Next, find the added volumes of the combined volumes. Finally, divide the moles by the final liters, and find the new molarity.

Q8

Determine concentration of all ions present in the solution when 3.0 g of calcium nitrate is dissolved in water to produce 15.0 mL of solution.

A4.30P

\([Ca^{2+}] = 1.2\;M\) and \([NO_3^-] = 2.4\;M\)

H4.30P

Need to be able to convert mole or gram into molarity in the given volume of solution
http://chemwiki.ucdavis.edu/Textbook_Maps/Map%3A_Zumdahl_8e/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.07_Stoichiometry_of_Precipitation_Reactions

S4.30P

The molar mass of \(Ca(NO_3)_2 = 164.088 \;g/mol\).

\[[Ca^{2+}] = \dfrac{\dfrac{ 3.0\; g }{ 164.088\; g/mole} }{ 15.0 \times 10^{-3} \;L } = 1.2\;M\]

Since the stoichiometry between calcium and nitrate is 1:2,

\[[NO_3^-] = 2.4\;M\]