reason, the NaOH coefficient cannot be 1. reason, the NaOH coefficient cannot be 3. reason, the NaOH coefficient cannot be 4.

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1 College Chemistry Problem Drill 11: Balancing Equations No. 1 of Inspection method of balancing equation is the basic approach used mostly for simple reactions. It all starts out by picking the first atom to balance, usually the one in the most complicated compound (i.e. with most atoms in a formula). When the following equation is balanced, what is the coefficient for NaOH? H 2 CO 3 (aq) + NaOH (aq) Na 2 CO 3 (aq) + H 2 O (l) (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 The most complex species is Na 2 CO 3. Try to balance the Na atom first. For that reason, the NaOH coefficient cannot be 1. Good job! You have balanced the equation correctly! Follow the three steps (1) Pick an atom Na, from the most complex species; (2) Balance the rest; (3) Verify each element and charge. The most complex species is Na 2 CO 3. Try to balance the Na atom first. For that reason, the NaOH coefficient cannot be 3. The most complex species is Na 2 CO 3. Try to balance the Na atom first. For that reason, the NaOH coefficient cannot be 4. The most complex species is Na 2 CO 3. Try to balance the Na atom first. For that reason, the NaOH coefficient cannot be 5. Use the Inspection Method to balance this simple equation: 1. Pick an atom from the most complex species Na 2 CO 3. In this case, start with the atom Na. To balance the Na, add 2 on the NaOH at the reactant side. 2. Balance the H and OH Just add 2 on the H 2 O. 3. Verify each element to see if all are balanced. We then get: 1 H 2 CO NaOH 1 Na 2 CO H 2 O Answer: (B) 2

2 No. 2 of One method to balance an equation with single element species (i.e. O 2 ) is to use fractions during the balancing process and then convert them into whole number coefficients at the end. For a reaction H 2 + O 2 H 2 O, you can balance it by adding a fraction ½ for O 2, i.e. H 2 + ½O 2 H 2 O (balanced) and then remove the fraction by multiplying 2 on both sides: 2H 2 + O 2 2H 2 O. When the following equation is balanced, what is the coefficient for O 2? KClO 3 KCl + O 2 (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Use the inspection method to balance the O with a fraction (3/2). B. Incorrect. Use the inspection method to balance the O with a fraction (3/2). C. Correct! Good job! This can be balanced by the Inspection Method. Simply check both ends and use a fraction to balance the only unbalanced element O. Remove the fraction and verify the equation. Use the inspection method to balance the O with a fraction (3/2). Use the inspection method to balance the O with a fraction (3/2). This can be worked out by the simple inspection method. (1) Check both ends, only O is not balanced. Use the fraction to balance the only unbalanced element. Add 3/2 on the O 2. (2) Remove the fraction by multiple 2 on each side (every species). (3) Verify each atom to see if all are balanced. We then get: 2 KClO 3 2 KCl + 3 O 2 Answer: (C) 3

3 No. 3 of The basic principle of the Oxidation Number method of balancing equations is that the number of electrons lost equals to the number of electron gained, i.e. charge conservation. Use this method to balance the redox equation: K 2 Cr 2 O 7 + HCl CrCl 3 + Cl 2 + H 2 O + KCl. What is the coefficient on HCl? (A) 10 (B) 14 (C) 7 (D) 5 (E) 2 Follow the four-step process of Oxidation Number Method. Assign the oxidation numbers and the net increase (Cl)/decrease (Cr) in charge. Multiply the proper ratio to equal the oxidized to the reduced. Finish the balancing with inspection. Good job! Follow the four-step process of Oxidation Number Method. Assign the oxidation numbers and the net increase (Cl)/decrease (Cr) in charge. Multiply the proper ratio to equal the oxidized to the reduced. Finish the balancing with inspection. Follow the four-step process of Oxidation Number Method. Assign the oxidation numbers and the net increase (Cl)/decrease (Cr) in charge. Multiply the proper ratio to equal the oxidized to the reduced. Finish the balancing with inspection. Follow the four-step process of Oxidation Number Method. Assign the oxidation numbers and the net increase (Cl)/decrease (Cr) in charge. Multiply the proper ratio to equal the oxidized to the reduced. Finish the balancing with inspection. Follow the four-step process of Oxidation Number Method. Assign the oxidation numbers and the net increase (Cl)/decrease (Cr) in charge. Multiply the proper ratio to equal the oxidized to the reduced. Finish the balancing with inspection. Step 1 and Step 2 (balance the redox atoms first): K 2 Cr 2 O 7 + 2HCl 2CrCl 3 + Cl 2 + H 2 O + KCl Being oxidized: Cl - to Cl 0 Δ =+1 or 2Cl - to Cl 2, the net change = 2x(+1)=+2 Being reduced: Cr 6+ to Cr 3+ Δ = -3 or Cr 2 6+ to 2Cr 3+, the net change = 2x(-3)=-6 Step 3: Multiply the oxidized (Cl) x3 and the reduced (Cr) x1 by a ratio of 3:1 to cancel out the change in charges. K 2 Cr 2 O 7 + 3x2HCl 2CrCl 3 + 3Cl 2 + H 2 O + KCl Step 4: Balance the rest with Inspection Method for K, H and O, finally Cl. K 2 Cr 2 O HCl 2CrCl 3 + 3Cl 2 + 7H 2 O + 2KCl Therefore the coefficient for HCl is 14. The correct answer is (B).

4 No. 4 of Butane (C 4 H 10 ) is an alkane used for liquefied petroleum gas (LPG) and butane lighter. The combustion reaction of butane with O 2 yields CO 2 and H 2 O. After balancing this equation, determine the coefficient for O 2. Hint: Balance the element in multiple reactants or products last. For combustion, balance O 2 last, with a fraction if needed. (A) 3 (B) 13 (C) 4 (D) 5 (E) 13/2 Pick an atom (C) from the most complex compound (C 4 H 10 ) to start. Continue on to balance the second atom (H). Use the fraction to balance the last atom (O). Finally remove the fraction to make each coefficient a whole number. Verify atom counts. Good job! Pick an atom (C) from the most complex compound (C 4 H 10 ) to start. Continue on to balance the second atom (H). Use the fraction to balance the last atom (O). Remove the fraction to make each coefficient a whole number. Verify atom counts. Pick an atom (C) from the most complex compound (C 4 H 10 ) to start. Continue on to balance the second atom (H). Use the fraction to balance the last atom (O). Finally remove the fraction to make each coefficient a whole number. Verify atom counts. Pick an atom (C) from the most complex compound (C 4 H 10 ) to start. Continue on to balance the second atom (H). Use the fraction to balance the last atom (O). Finally remove the fraction to make each coefficient a whole number. Verify atom counts. The final balanced equation should have whole numbers as coefficients, not fractions. Pick an atom (C) from the most complex compound (C 4 H 10 ) to start. Continue on to balance the second atom (H). Use the fraction to balance the last atom (O). Finally remove the fraction to make each coefficient a whole number. Verify the atom counts. Write out the reaction: C 2 H 10 + O 2 CO 2 + H 2 O Use the Inspection Method with Fraction: #1. Start out by picking the first atom from the most complicated species. In this case, pick C from C 4 H 10. Conveniently, you can use the table to tabulate the coefficients. C 4 H 10 + O 2 4CO 2 + H 2 O #2. Next to balance the second element H from C 4 H 10. C 4 H 10 + O 2 4CO 2 + 5H 2 O #3. Balance the last element O using a fraction. C 4 H 10 + (13/2)O 2 4CO 2 + 5H 2 O #4. Lastly, remove the fraction by multiply by 2 on both sides. 2C 4 H O 2 8CO H 2 O Verify the atom counts for each element on both sides. Answer: (B) 13

5 No. 5 of For ionic reactions, after balancing the mass, you need to make certain that the total charges are the same on both sides. What are the coefficients of the following equation when it is balanced? Mg + Na + Na + Mg 2+ (A) Mg + Na + Na + Mg 2+ (B) Mg + 2Na + 2Na + Mg 2+ (C) 2Mg + Na + Na + 2Mg 2+ (D) 2Mg + 2Na + 2Na + 2Mg 2+ (E) Mg + 2Na + Na + Mg 2+ Each element has been balanced. Make sure to balance the charges as well. Good job! The atoms are balanced as is. Use Na to balance the charge. Each element has been balanced. Make sure to balance the charges as well. Each element has been balanced. Make sure to balance the charges as well. The charges are balanced but not the elements. Remember to balance charges as well as atoms. This is really just a one-step act. Check both sides, the atoms have been balanced. The charge must also be balanced. Just use Na to balance the charge. 1 Mg + 2 Na + 2 Na + 1 Mg 2+ Answer: (B) Mg + 2Na + 2Na + Mg 2+

6 No. 6 of In balancing an equation, do not change the subscripts. This could result in a totally different compound. To balance, it is all about adjusting the coefficients, which show how many molecules of a specific compound involves in a reaction. Subscripts show how many atoms of the same element in one molecule. Write and balance the following double displacement reaction in an aqueous solution: CaCl 2 + Na 2 SO 4? (A) CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + Na 2 Cl 2 (aq) (B) CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (aq) + 2NaCl (aq) (C) CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + 2NaCl (aq) (D) CaCl 2 (aq) + Na 2 SO 4 (aq) 2CaSO 4 (s) + NaCl (aq) (E) 2CaCl 2 (aq) + 2Na 2 SO 4 (aq) 2CaSO 4 (s) + 4NaCl (aq) Do not change the subscripts of any formula in reactants or products. Na 2 Cl 2 is not the correct formula for sodium chloride. B. Incorrect. The reaction is balanced correctly; however, the CaSO 4 is a precipitate which should be denoted as a solid (s). C. Correct. CaSO 4 is the solid product. The equation is correctly balanced with equal atoms on both sides. The equation is not balanced. Since there are two Cl s and Na s at the reactant side, just double the coefficient for the product NaCl to balance the equation. Although the equation is balanced with equal number of atoms at both sides, the coefficients should be in their lowest terms. Divide each by 2. You will get the lowest coefficients for all. A double displacement reaction is a type of reaction where cations and anions of two different compounds switch places. To balance these reactions, treat each polyatomic cation or anion as a whole unit. Balance them as a group, instead of single atom. For example, SO 4 is a polyatomic ion and it appears at both sides unbroken due to the nature of double displacement. Just count SO 4 as a whole group. Since there is one SO 4 on each side, SO 4 is balanced. This equation can be balanced in one single step. Since there are two Cl s and two Na s at the reactant side, just add the coefficient of 2 for NaCl. According to the solubility rule, CaSO 4 is a precipitate. Answer: (C) CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s)+ 2NaCl (aq)

7 No. 7 of To balance a redox reaction in basic solution, use 2OH - /H 2 O to add O and H 2 O/OH - to add H to the side needing it. Balance the following ionic redox in alkaline solution using half-reaction method and determine the coefficient in MnO MnO 4 MnO O 2 (A) 2 (B) 3 (C) 4 (D) 5 (E) 1 Follow the 9-step process to balance a redox in basic solution. Write out each half reaction with presence of OH -. Add OH - and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. B. Incorrect. Follow the 9-step process to balance a redox in basic solution. Write out each half reaction with presence of OH -. Add OH - and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. C. Correct. Good job! Follow the 9-step process to balance a redox in basic solution. Write out each half reaction with presence of OH -. Add OH - and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Follow the 9-step process to balance a redox in basic solution. Write out each half reaction with presence of OH -. Add OH - and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Follow the 9-step process to balance a redox in basic solution. Write out each half reaction with presence of OH -. Add OH - and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Reduced: Mn (+7) O 4 - to Mn (+6) O 4 2-, Δ=-1. MnO e MnO 4 2- Oxidized (with the presence of OH - ): 2O (-2) H - to O 2 (0). Δ=2x(+2)=+4. 2OH - O 2 + 2e - ; Use OH - /H 2 O to add H on the product side and e - to balance the charge. 4OH - O 2 + 2H 2 O + 4e - Multiply factors to cancel the electrons e -. 4x [MnO e MnO 2-4 ] 1x [4OH - O 2 + 2H 2 O + 4e - ] Add them together and cancel the 4e - : 4OH MnO 4 4MnO 4 + O 2 + 2H 2 O (balanced) The correct answer is (C).

8 No. 8 of One useful technique to balance double displacement reactions is to treat a polyatomic ion as a group and count it as a larger entity, instead of atom by atom, since such a group is intact during the reaction. Let s put this into practice. Write out and balance the double displacement reaction CaCO 3 + H 3 PO 4? (A) 3CaCO 3 + 2H 3 PO 4 Ca 3 (PO 4 ) 2 + 3H 2 CO 3 (B) 3CaCO 3 + 2H 3 PO 4 Ca 3 (PO 4 ) 2 + 3H 2 O + 3CO 2 (C) 3CaCO 3 + 3H 3 PO 4 3CaPO 4 + 3H 2 O + 3CO 2 (D) 6CaCO 3 + 4H 3 PO 4 2Ca 3 (PO 4 ) 2 + 6H 2 O + 6CO 2 (E) CaCO 3 + (2/3)H 3 PO 4 (1/3)Ca 3 (PO 4 ) 2 + H 2 CO 3 First write out the products. Since this is the double displacement, their anions and cations are exchanged. The products should be Ca 3 (PO 4 ) 2 + H 2 CO 3, but H 2 CO 3 will decompose into CO 2 and H 2 O. To balance it, pick PO 4 first as a group and then next pick Ca. The rest will come easily. Good job! Since this is the double displacement, their anions and cations are exchanged. The products should be Ca 3 (PO 4 ) 2 + H 2 CO 3, but H 2 CO 3 will decompose into CO 2 and H 2 O. Treat phosphate as a group to balance first. Everything else becomes easy. Calcium phosphate is Ca 3 (PO 4 ) 2 not CaPO 4. If the wrong product is written, balancing becomes a lot harder. That is one indication that something is wrong about the approach. The equation is balanced but all coefficients need to be simplified. Divide each by 2. The equation is balanced but fractional coefficients are used. You should convert any fraction into an integer. In this case, multiply each side by 3. Since this is a double displacement reaction, exchange their anions and cations. Carbonic acid yields to CO 2 and H 2 O at a normal condition. Therefore the reaction is as follows: CaCO 3 + H 3 PO 4 Ca 3 (PO 4 ) 2 + H 2 O + CO 2 Now to balance above equation: #1: Pick the polyatomic group PO 4 to balance first. Treat PO 4 group as one entity. Write out 1 on the most complex molecule. Balance this group on the other side. CaCO 3 + 2H 3 PO 4 -> 1Ca 3 (PO 4 ) 2 + H 2 O + CO 2 #2: Pick Ca to balance next. 3CaCO 3 + 2H 3 PO 4 -> 1Ca 3 (PO 4 ) 2 + H 2 O + CO 2 #3: Then balance C, H and O. The last element is O. We can omit oxygen atoms in PO 4 group as they are already balanced. 3CaCO 3 + 2H 3 PO 4 -> 1Ca 3 (PO 4 ) 2 + 3H 2 O + 3CO 2 Check and verify each atom count. Answer: (B) 3CaCO 3 + 2H 3 PO 4 Ca 3 (PO 4 ) 2 + 3H 2 O + 3CO 2

9 No. 9 of Half reaction method is to separate the overall redox into two half reactions oxidation and reduction. In acidic solution, balance the oxygen atoms by adding water and balance hydrogen atoms by adding hydrogen ions (from acidic solution). Balance the net ion equation for a reaction below in acid solution using the half reaction method. What is the coefficient for Pb 2+? MnO + PbO 2 MnO Pb 2+ (A) 5 (B) 4 (C) 3 (D) 2 (E) 1 A. Correct. Good job! Follow the 9-step process to balance a redox in acidic solution. Write out each half reaction. Add H + and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. B. Incorrect. Follow the 9-step process to balance a redox in acidic solution. Write out each half reaction. Add H + and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Follow the 9-step process to balance a redox in acidic solution. Write out each half reaction. Add H + and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Follow the 9-step process to balance a redox in acidic solution. Write out each half reaction. Add H + and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Follow the 9-step process to balance a redox in acidic solution. Write out each half reaction. Add H + and H 2 O as needed to balance H and O. Make sure both elements and charges are balanced. Reduced: Mn 2+ to Mn 7+ and Oxidized: Pb 4+ to Pb 2+. Split two half-reactions: Mn 2+ MnO e - and PbO 2 + 2e Pb 2+ Balance each with H + and H 2 O: MnO + 3H 2 O MnO e - + 6H + PbO 2 + 2e - + 4H + Pb H 2 O Multiply factors to cancel the electrons e -. 2x [MnO + 3H 2 O MnO e - + 6H + ] 5x [PbO 2 + 2e - + 4H + Pb H 2 O] Add them together and cancel the like terms: 2MnO + 5PbO 2 + 8H + 2MnO Pb H 2 O (balanced) The correct answer is (A).

10 No. 10 of The key in balancing an equation is to pick the first atom to start an atom from the most complex molecule. The second atom to balance should also be picked from the same molecule. Write down 1 on the most complex molecule first and start the process. Use fractions if needed. Write out and balance the ethylene glycol combustion C 2 H 6 O 2 + O 2. (A) C 2 H 6 O 2 + (5/2)O 2 2CO 2 + 3H 2 O (B) 2C 2 H 6 O 2 + 5O 2 4CO 2 + 6H 2 O (C) C 2 H 6 O 2 + 4O 2 2CO 2 + 6H 2 O (D) C 2 H 6 O 2 + 5O 2 4CO 2 + 6H 2 O (E) 2C 2 H 6 O 2 + 5O 2 4CO 2 + 4H 2 O The equation is balanced but a fraction is used. Use the Inspection Method to start on C first from the most complex molecule C 2 H 6 O 2, then H. For combustion, the last element to balance is O. Use fractions to balance the oxygen as needed. Remove the fractions at the final step. Good job! Use the Inspection Method to start on C first from the most complex molecule C 2 H 6 O 2, then H. For combustion, the last element to balance is O. Use fractions to balance the oxygen as needed. Remove the fractions at the final step. Use the Inspection Method to start on C first from the most complex molecule C 2 H 6 O 2, then H. For combustion, the last element to balance is O. Use fractions to balance the oxygen as needed. Remove the fractions at the final step. Use the Inspection Method to start on C first from the most complex molecule C 2 H 6 O 2, then H. For combustion, the last element to balance is O. Use fractions to balance the oxygen as needed. Remove the fractions at the final step. Use the Inspection Method to start on C first from the most complex molecule C 2 H 6 O 2, then H. For combustion, the last element to balance is O. Use fractions to balance the oxygen as needed. Remove the fractions at the final step. This is a typical combustion reaction with the products of CO 2 and H 2 O. C 2 H 6 O 2 + O 2 CO 2 + H 2 O Use the Inspection Method with Fraction: #1. Pick C to start from the most complex molecule C 2 H 6 O 2. Place 1 on C 2 H 6 O 2. 1C 2 H 6 O 2 + O 2 -> 2CO 2 + H 2 O #2. Continue onto the next, H, the second element from C 2 H 6 O 2. 1C 2 H 6 O 2 + O 2 2CO 2 + 3H 2 O #3. O is the last to balance since this is a combustion reaction. It also means that you should consider using a fraction to balance the oxygen. 1C 2 H 6 O 2 + (5/2)O 2 2CO 2 + 3H 2 O #4. Remove the fraction by multiplying 2 on both sides. 2C 2 H 6 O 2 + 5O 2 4CO 2 + 6H 2 O Answer: (B) 2C 2 H 6 O 2 + 5O 2 4CO 2 + 6H 2 O