Solution Set 2

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4.1

The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia, which is important in the production of fertilizer. The unbalanced chemical reaction is

$N_2(g)+H_2(g) \rightleftharpoons NH_3(g)$

Under certain conditions, 50.0 grams of nitrogen are converted to ammonia, and the process has a 66.0% yield of ammonia. What mass of ammonia is produced?

Answer $$40.0\; g$$

Hint: Balance the equation and then calculate the moles of ammonia by stoichiometry. After get the theoretical mass of the ammonia, use the percent yield formula to get the actual mass. 3.4 Mass Relationships in Chemical Equations

Solution: first write out the balanced chemical equation:

$N_{2\; (g)}+3H_{2\;(g)} \rightleftharpoons 2NH_{3\;(g)}$

• moles of $$N_2$$: 50.0g/(14.01x2)g/mol= 1.78moles, by stoichiometry,
• moles of $$NH_3$$: (2mol NH3/1mol N2)x1.78moles N2= 3.56 moles NH3
• mass of $$NH_3$$ : (theoretical)= 3.56 moles NH3 x (14.01+1.008 x 3) g/mole = 60.6grams
• percent yield: (actual/ theoretical) x 100%, actual = 66.0%x60.6g= 40.0grams

4.2

Vinegar, an aqueous solution of acetic acid, $$HC_2H_3O_2$$, can be used to remove the calcium carbonate scale ($$CaCO_3$$) left on pots and pans when hard water has been used for cooking. If you use 65.5 g of acetic acid, how many moles of calcium carbonate can you dissolve?

$HC_2H_3O_2 + CaCO_3 \rightarrow Ca(C_2H_3O_2) + CO_2 + H_2O \; \rm{(unbalanced)}$

Answer 0.545 mol $$CaCO_3$$

Hint: To compare different elements, a chemical equation must be balanced. See http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions

Solution: First balance the chemical equation. Check that the same number of atoms of each element appears on both sides of the arrow.

$\underline{2} HC_2H_3O_2 + CaCO_3 \rightarrow Ca(C_2H_3O_2) + CO_2 + H_2O$

Then create a blank template to check that units cancel out until what’s left is what the question is asking for. Start out with the information known ($$g$$ of acetic acid) and use molecular weights and the formula to convert into mol $$CaCO_3$$.

g HC2H3O2 * (mol/g HC2H3O2) * (mol CaCO3/mol HC2H3O2) = mol CaCO3

Finally, plug in values and calculate the final answer. The molecular weight of acetic acid can be calculated using values from the periodic table, while the ratio between moles of calcium carbonate and moles of acetic acid can be found from the formula. The answer should be rounded to three significant figures.

65.5 g HC2H3O2 * (1 mol/60.0524 g HC2H3O2) * (1mol CaCO3/2mol HC2H3O2) = 0.545 mol CaCO3

4.3

Aluminum and sulfur react to form aluminum sulfide.

$2 Al + 3 S \rightarrow Al_2S_3$

1. If 6.75 g aluminum and 9.25 g sulfur are mixed, how much aluminum sulfide can be produced?
2. Which react is the limiting reagent?
3. How much of the other reagent is left after the reaction is complete?

1. 14.4 g $$Al_2S_3$$
2. Sulfur, S
3. 1.56 g Al

Hint: The limiting reagent determines how much of each product can be produced. See http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Limiting_Reagents

Solution:

1. First make an assumption on which is the limiting reagent. Check the assumption by calculating how much of the other reagent is needed to react—first create a blank template to make sure units cancel out into g of the assumed excess reagent, then plug in values. Use the periodic table to calculate molecular weights and the chemical equation to convert from moles of one substance to moles of another substance.

Assuming aluminum is the limiting reagent:

6.75 g Al * (1 mol Al/26.9815 g Al) * (3 mol S/2 mol Al) * (32.06 g S/1 mol S)=12.0 g S

Since 9.25 g S < 12.0 g S, sulfur must be the limiting reagent.

Next, start with the mass of the limiting reagent and convert into mass of the product. Follow the steps used in the previous calculation.

9.25 g S * (1 mol S/32.06 g S) * (1 mol Al2S3/3 mol S) * (150.143 g/mol Al2S3)

= 14.4 g Al2S3

1. As found in part (a), the limiting reagent is sulfur.
2. As in part (a), calculate how much aluminum is needed to react with the given mass of sulfur. Then subtract that value from the given mass of aluminum to calculate how much aluminum is left after completion of the reaction.

9.25 g S * (1 mol S/32.06 g S) * (2 mol Al/3 mol S) * (26.9815 g/mol Al) = 5.19 g Al

6.75 g Al – 5.19 g Al = 1.56 g Al left

4.4

Acetylsalicylic acid ($$C_9H_8O_4$$), or aspirin, can be made from acetic anhydride and salicylic acid.

$C_7H­_6O_3 + C_4H_6O_3 \rightarrow C_9H_8O_4 + C_2H_4O_2$

Salicylic acid + acetic anhydride → acetylsalicylic acid + acetic acid

How much C9H8O4 and C2H4O2 will be produced if 5.32 g of $$C_7H­_6O_3$$ reacts with 5.11 g of $$C_4H_6O_3$$? Assume 100% yield.

Answer: 6.94 g $$C_9H_8O_4$$; 2.31 g $$C_2H_4O_2$$

Hint: The limiting reagent determines how much of each product can be produced. See http://chemwiki.ucdavis.edu/Analytic...iting_Reagents

Solution: First, determine the limiting reagent. Make an assumption on which is the limiting reagent. Check the assumption by calculating how much of the other reagent is needed to react—first create a blank template to make sure units cancel out into g of the assumed excess reagent, then plug in values. Use the periodic table to calculate molecular weights and the chemical equation to convert from moles of one substance to moles of another substance.

Assuming $$C_7H­_6O_3$$ is the limiting reagent:

5.32 g C76O3 * (1 mol/138.1172 g C76O3) * (1 mol C4H6O3/1 mol C76O3) * (102.0842 g/mol C4H6O3) = 3.93 g C4H6O3

Because the problem states that there is 5.11 g C4H6O3, the limiting reagent is C76O3.

To calculate the amount of each product produced, start with the mass of the limiting reagent and convert to g of desired substance. Since there is a 1:1:1:1 ratio in the equation, the only difference in calculating the mass of each product is the molecular weights of the compounds.

• 5.32 g C76O3 * (1 mol/138.1172 g C76O3) * (1 mol C9H8O4/1 mol C76O3) * (180.1526 g/mol C9H8O4) = 6.94 g C9H8O4
• 5.32 g C76O3 * (1 mol/138.1172 g C76O3) * (1 mol C2H4O2/1 mol C76O3) * (60.0488 g/mol C2H4O2) = 2.31 g C2H4O2

4.5

Baking soda (sodium bicarbonate) has a variety of uses, including cleaning, dental care (toothpastes), cooking, and neutralizing acid spills on highways. When it reacts with sulfuric acid, it produces carbon dioxide.

$2 NaHCO_3 + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O + 2 CO_2$

A chemistry student reacts 375 g $$NaHCO_3$$ with 225 g $$H_2SO_4$$.

1. How much $$CO_2$$ forms? (Assume 100% yield)
2. Identify the limiting and excess reagents.
3. How much excess reagent is left after the reaction is complete?
4. What is the percent yield if 150 g of $$CO_2$$ is produced?

1. 196 g CO­2
2. limiting: NaHCO3; excess: H2SO4
3. 6 g H2SO4
4. 76.5%

Hint: The limiting reagent determines how much of each product can be produced. See http://chemwiki.ucdavis.edu/Analytic...iting_Reagents

Solution:

1. First, determine the limiting reagent. Make an assumption on which is the limiting reagent. Check the assumption by calculating how much of the other reagent is needed to react—first create a blank template to make sure units cancel out into g of the assumed excess reagent, then plug in values. Use the periodic table to calculate molecular weights and the chemical equation to convert from moles of one substance to moles of another substance.

Assuming NaHCO3 is the limiting reagent:

375 g NaHCO3 * (1 mol/84.0069 g NaHCO3) * (1 mol H2SO4/2 mol NaHCO3) * (98.0734 g/mol H2SO4) = 219 g H2SO4

Since the student has 225 g H2SO4, the limiting reagent is NaHCO3.

Now convert the original mass of NaHCO3 into g of the desired product. Follow the same steps to solve the previous calculation.

375 g NaHCO3* (1 mol/84.0069 g NaHCO3) * (2 mol CO2/2 mol NaHCO3) * (44.0098 g/mol CO2) = 196 g CO2

1. From part (a), the limiting reagent was determined to be NaHCO3 and H2SO4 is the excess reagent.
2. Subtract the value of H2SO4 that reacts with the NaHCO3 (this was calculated in part (a)) from the original mass of H2SO4 to calculate how much H2SO remains.

225 g – 219 g = 6 g H2SO4

1. Percent yield = (actual yield/theoretical yield) * 100%. The actual yield is given as 150. g and the theoretical yield was calculated in part (a). Plug in the numbers and solve.

(150. g/196 g) * 100% = 76.5%

4.6

Para-dichlorobenzene, which is frequently used to repel moths as mothballs, has a percentage composition of 49.0% $$C$$, 2.7% $$H$$, and 48.3% $$Cl$$. It has a molecular weight of 147.0 g/mol. Find its empirical and molecular formulas.

Answer Empirical formula is $$C_3H_2Cl$$ and Molecular formula is $$C_6H_4Cl_2$$

Hint: Percent composition shows the percent of the molecular mass that each element composes. See Determining Empirical and Molecular Formulas

Solution: Multiply the molecular weight by the decimal representation of each percent to get the masses of each element in the compound.

$147.0\; g \times 0.490 = 72.0\; g\; \text{Carbon}$

$147.0\; g \times 0.027 = 4.0\; g\; \text{Hydrogen}$

$147.0\; g \times 0.483 = 71.0\; g\; \text{Chlorine}$

Calculate the moles of each substance using atomic weights from the periodic table.

72.0 g C * (1 mol/12.011 g C) = 5.99 mol C

4.0 g H * (1 mol/1.0079 g H) = 4.0 mol H

71.0 g Cl * (1 mol/35.453 g Cl) = 2.00 mol Cl

Divide each number by the smallest number of moles; the empirical formula is then written based off the whole-number ratios of moles.

5.99 mol/2.00 mol = 3

4.0 mol/2.00 mol = 2

2.00 mol/2.00 mol = 1

Empirical formula: $$C_3H_2Cl$$

To find the molecular formula, divide the molecular weight by the empirical formula weight. Then multiply the subscript of each element in the empirical formula by the determined factor to get the molecular formula.

M.W. C3H2Cl = 73.5018 g/mol

$\dfrac{147.0\; g}{73.5018\; g} = 2$

Molecular formula: $$C_{3 \times 2}H_{2 \times 2}Cl_{1 \times 2} = C_6H_4Cl_2$$

4.7

Balance the following reactions.

1. $$NaHCO_{3\;(s)} \rightarrow Na_2CO_{3\;(s)} + CO_{2\;(g)} + H_2O_{(g)}$$
2. $$FeSO_{4\;(aq)} + Ag_{(s)} \rightarrow Ag_2SO_{4\;(s)} + Fe_{(s)}$$
3. $$CaCl_{2\;(aq)} + K_2CO_{3\;(aq)} \rightarrow CaCO_3(s) + KCl_{(aq)}$$
4. $$C_2H_5OH_{(l)} + O_{2\;(g)} \rightarrow CO_2(g) + H_2O(l)$$
5. $$FeCl_{3\;(aq)} + NaOH_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + NaCl_{(aq})$$
6. $$CS_{2\;(s)} + Cl_{2\;(g)} \rightarrow CCl_4(l) + SCl_2(s)$$
7. $$Na_3PO_{4\;(aq)} + CrCl_{3\;(aq)} \rightarrow CrPO_{4\;(s)} + NaCl_{(aq)}$$

1. $$\underline{2} NaHCO_{3\;(s)} \rightarrow Na_2CO_{3\;(s)} + CO_{2\;(g)} + H_2O_{(g)}$$
2. $$FeSO_{4\;(aq)} + \underline{2} Ag_{(s)} \rightarrow Ag_2SO_{4\;(s)} + Fe_{(s)}$$
3. $$CaCl_{2\;(aq)} + K_2CO_{3\;(aq)} \rightarrow CaCO_3(s) + \underline{2} KCl_{(aq)}$$
4. $$C_2H_5OH_{(l)} + \underline{3} O_{2\;(g)} \rightarrow \underline{2}CO_2(g) +\underline{3} H_2O(l)$$
5. $$FeCl_{3\;(aq)} + \underline{3} NaOH_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + \underline{3} NaCl_{(aq})$$
6. $$CS_{2\;(s)} + \underline{4}Cl_{2\;(g)} \rightarrow CCl_4(l) + \underline{2} SCl_2(s)$$
7. $$Na_3PO_{4\;(aq)} + CrCl_{3\;(aq)} \rightarrow CrPO_{4\;(s)} + \underline{3} NaCl_{(aq)}$$

Solution:

1. _ NaHCO3(s) → _ Na2CO3(s) + _ CO2(g) + _ H2O(g)
• Balance Na atoms: 2 NaHCO3(s) → 1 Na2CO3(s) + CO2(g) + H2O(g)
• All atoms are balanced, 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
1. _ FeSO4(aq) + _ Ag(s) → _ Ag2SO4(s) + _ Fe(s)
• Fe atom is balanced, 1
• Balance Ag atoms: 1
• All atoms are balanced, FeSO4(aq) + 2 Ag(s) → Ag2SO4(s) + Fe(s)
1. _ CaCl2(aq) + _ K2CO3(aq) → _ CaCO3(s) + _ KCl(aq)
• Ca is balanced, 1 CaCl2(aq) + _ K2CO3(aq) 1 CaCO3(s) + _ KCl(aq)
• Balance Cl atoms: 1 CaCl2(aq) + _ K2CO3(aq) 1 CaCO3(s) + 2 KCl(aq)
• All atoms are balanced, CaCl2(aq) + K2CO3(aq) CaCO3(s) + 2 KCl(aq)
1. _ C2H5OH(l) + _ O2(g) → _ CO2(g) + _ H2O(l)
• Balance C atoms: 1 C2H5OH(l) + _ O2(g) → 2 CO2(g) + _ H2O(l)
• Balance H atoms: 1 C2H5OH(l) + _ O2(g) → 2 CO2(g) + 3 H2O(l)
• Balance O atoms: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
1. _ FeCl3(aq) + _ NaOH(aq) → _ Fe(OH)3(s) + _ NaCl(aq)
• Fe atom is balanced, 1 FeCl3(aq) + _ NaOH(aq) → 1 Fe(OH)3(s) + _ NaCl(aq)
• Balance Cl atoms: 1 FeCl3(aq) + _ NaOH(aq) → 1 Fe(OH)3(s) + 3 NaCl(aq)
• Balance Na atoms: 1 FeCl3(aq) + 3 NaOH(aq) → 1 Fe(OH)3(s) + 3 NaCl(aq)
• All atoms are balanced, FeCl3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaCl(aq)
1. _ CS2(s) + _ Cl2(g) → _ CCl4(l) + _ SCl2(s)
• C atom is balanced, 1 CS2(s) + _ Cl2(g) 1 CCl4(l) + _ SCl2(s)
• Balance S atoms: 1 CS2(s) + _ Cl2(g) 1 CCl4(l) + 2 SCl2(s)
• Balance Cl atoms: CS2(s) + 4 Cl2(g) CCl4(l) + 2 SCl2(s)
1. _ Na3PO4(aq) + _ CrCl3(aq) → _ CrPO4(s) + _ NaCl(aq)
• Balance Na atoms: 1 Na3PO4(aq) + _ CrCl3(aq) _ CrPO4(s) + 3 NaCl(aq)
• All atoms are balanced, Na3PO4(aq) + CrCl3(aq) CrPO4(s) + 3 NaCl(aq)

4.8

Give the balanced equation for each of the following.

1. $$HClO_{4\;(aq)} + Mg(OH)_{2\;(s)} \rightarrow Mg(ClO_4)_{2\;(aq)} + H_2O_{(l)}$$
2. $$H_2SO_{4\;(aq)} + Pb(NO_3)_{2\;(aq)} \rightarrow PbSO_{4\;(s)} + HNO_{3\;(aq)}$$
3. $$NaOH_{(aq)} + AlCl_{3\;(aq)} \rightarrow Al(OH)_{3\;(s)} + NaCl_{(aq)}$$

1. $$\underline{2} HClO_{4\;(aq)} + Mg(OH)_{2\;(s)} \rightarrow Mg(ClO_4)_{2\;(aq)} + \underline{2}H_2O_{(l)}$$
2. $$H_2SO_{4\;(aq)} + Pb(NO_3)_{2\;(aq)} \rightarrow PbSO_{4\;(s)} + \underline{2} HNO_{3\;(aq)}$$
3. $$\underline{3}NaOH_{(aq)} + AlCl_{3\;(aq)} \rightarrow Al(OH)_{3\;(s)} + \underline{3}NaCl_{(aq)}$$

Solution:

1. _ HClO4(aq) + _ Mg(OH)2(s) → _ Mg(ClO4)2(aq) + _ H2O(l)
• Balance Cl atoms: 2 HClO4(aq) + _ Mg(OH)2(s) 1 Mg(ClO4)2(aq) + _ H2O(l)
• Balance Mg atoms: 2 HClO4(aq) + 1 Mg(OH)2(s) 1 Mg(ClO4)2(aq) + _ H2O(l)
• Balance H and O atoms: 2 HClO4(aq) + Mg(OH)2(s) Mg(ClO4)2(aq) + 2 H2O(l)
1. _ H2SO4(aq) + _ Pb(NO3)2(aq) → _ PbSO4(s) + _ HNO3(aq)
• S atom is already balanced, 1 H2SO4(aq) + _ Pb(NO3)2(aq) → 1 PbSO4(s) + _ HNO3(aq)
• Balance N atoms: 1 H2SO4(aq) + 1 Pb(NO3)2(aq) → 1 PbSO4(s) + 2 HNO3(aq)
• All other atoms are balanced, H2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2 HNO3(aq)
1. _ NaOH(aq) + _ AlCl3(aq) → _ Al(OH)3(s) + _ NaCl (aq)
• Balance O atoms: 3 NaOH(aq) + _ AlCl3(aq) 1 Al(OH)3(s) + _ NaCl (aq)
• Balance Na atoms: 3 NaOH(aq) + _ AlCl3(aq) 1 Al(OH)3(s) + 3 NaCl (aq)
• Other atoms are balanced, 3 NaOH(aq) + AlCl3(aq) Al(OH)3(s) + 3 NaCl (aq)

4.9

Please give the balanced equation for each of the following.

1. $$Na_{(s)}+H_2O_{(l)} \rightarrow NaOH_{(aq)}+H_{2\;(g)}$$
2. $$H_{2\;(g)} +O_{2\;(g)} \rightarrow H_2O_{(g)}$$
3. $$AgNO_3 + CaCl_2 \rightarrow AgCl+Ca(NO_3)_2$$

1. $$\underline{2} Na_{(s)}+\underline{2} H_2O_{(l)} \rightarrow \underline{2} NaOH_{(aq)}+H_{2\;(g)}$$
2. $$\underline{2} H_{2\;(g)} +O_{2\;(g)} \rightarrow \underline{2} H_2O_{(g)}$$
3. $$\underline{2} AgNO_3 + CaCl_2 \rightarrow \underline{2} AgCl+Ca(NO_3)_2$$

Hint: Count the number of each element in both reactant and product. http://chemwiki.ucdavis.edu/Wikitexts/ChemTutor/Chemical_Reactions

Solution:

• H2O=OH+H, need one more H to form H2, 2H2O=2OH+H2. 2OH need 2Na to make 2NaOH.
• So 2Na+2H2O -> 2NaOH+H2