College Chemistry – Problem Drill 11: Balancing Equations

College Chemistry ? Problem Drill 11: Balancing Equations

Question No. 1 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.

1. 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?

H2CO3 (aq) + NaOH (aq) Na2CO3 (aq) + H2O (l)

Question

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

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A. Incorrect. The most complex species is Na2CO3. Try to balance the "Na" atom first. For that reason, the NaOH coefficient cannot be "1".

B. Correct. 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.

C. Incorrect. The most complex species is Na2CO3. Try to balance the "Na" atom first. For that reason, the NaOH coefficient cannot be "3".

D. Incorrect. The most complex species is Na2CO3. Try to balance the "Na" atom first. For that reason, the NaOH coefficient cannot be "4".

E. Incorrect. The most complex species is Na2CO3. Try to balance the "Na" atom first. For that reason, the NaOH coefficient cannot be "5".

Solution

Use the Inspection Method to balance this simple equation: 1. Pick an atom from the most complex species Na2CO3. 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 H2O. 3. Verify each element to see if all are balanced.

We then get: 1 H2CO3 + 2 NaOH 1 Na2CO3 + 2 H2O

Answer: (B) 2

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Question No. 2 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.

2. One method to balance an equation with single element species (i.e. O2) is to use fractions during the balancing process and then convert them into whole number coefficients at the end. For a reaction H2 + O2 H2O, you can balance it by adding a fraction ? for O2, i.e. H2 + ?O2 H2O (balanced) and then remove the fraction by multiplying 2 on both sides: 2H2 + O2 2H2O. When the following equation is balanced, what is the coefficient for O2?

KClO3 KCl + O2

Question

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

A. Incorrect. 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).

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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.

D. Incorrect. Use the inspection method to balance the "O" with a fraction (3/2).

E. Incorrect. 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 "O2". (2) Remove the fraction by multiple "2" on each side (every species).

(3) Verify each atom to see if all are balanced.

Solution

We then get: 2 KClO3 2 KCl + 3 O2

Answer: (C) 3

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Question No. 3 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.

3. 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: K2Cr2O7 + HCl CrCl3 + Cl2 + H2O + KCl. What is the coefficient on HCl?

Question

(A) 10 (B) 14 (C) 7 (D) 5 (E) 2

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A. Incorrect.

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.

B. Correct.

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.

C. Incorrect.

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.

D. Incorrect.

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.

E. Incorrect.

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.

Solution

Step 1 and Step 2 (balance the redox atoms first): K2Cr2O7 + 2HCl 2CrCl3 + Cl2 + H2O + KCl Being oxidized: Cl- to Cl0 =+1 or 2Cl- to Cl2, the net change = 2x(+1)=+2 Being reduced: Cr6+ to Cr3+ = -3 or Cr26+ to 2Cr3+, 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. K2Cr2O7 + 3x2HCl 2CrCl3 + 3Cl2 + H2O + KCl Step 4: Balance the rest with Inspection Method for K, H and O, finally Cl. K2Cr2O7 + 14HCl 2CrCl3 + 3Cl2 + 7H2O + 2KCl Therefore the coefficient for HCl is 14.

The correct answer is (B).

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Question No. 4 of 10

Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.

4. Butane (C4H10) is an alkane used for liquefied petroleum gas (LPG) and butane lighter. The combustion reaction of butane with O2 yields CO2 and H2O. After balancing this equation, determine the coefficient for O2. Hint: Balance the element in multiple reactants or products last. For combustion, balance O2 last, with a fraction if needed.

Question

(A) 3 (B) 13 (C) 4 (D) 5 (E) 13/2

A. Incorrect. Pick an atom (C) from the most complex compound (C4H10) 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.

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B. Correct. Good job! Pick an atom (C) from the most complex compound (C4H10) 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.

C. Incorrect. Pick an atom (C) from the most complex compound (C4H10) 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.

D. Incorrect. Pick an atom (C) from the most complex compound (C4H10) 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.

E. Incorrect. The final balanced equation should have whole numbers as coefficients, not fractions. Pick an atom (C) from the most complex compound (C4H10) 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.

Solution

Write out the reaction: C2H10 + O2 CO2 + H2O 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 C4H10. Conveniently, you can use the table to tabulate the coefficients. C4H10 + O2 4CO2 + H2O #2. Next to balance the second element "H" from C4H10. C4H10 + O2 4CO2 + 5H2O #3. Balance the last element "O" using a fraction. C4H10 + (13/2)O2 4CO2 + 5H2O #4. Lastly, remove the fraction by multiply by 2 on both sides. 2C4H10 + 13O2 8CO2 + 10H2O Verify the atom counts for each element on both sides.

Answer: (B) 13

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Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.

5. 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 + Mg2+

Question

(A) Mg + Na+ Na + Mg2+ (B) Mg + 2Na+ 2Na + Mg2+ (C) 2Mg + Na+ Na + 2Mg2+ (D) 2Mg + 2Na+ 2Na + 2Mg2+ (E) Mg + 2Na+ Na + Mg2+

A. Incorrect. Each element has been balanced. Make sure to balance the charges as well.

B. Correct. Good job! The atoms are balanced as is. Use "Na" to balance the charge.

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C. Incorrect. Each element has been balanced. Make sure to balance the charges as well.

D. Incorrect. Each element has been balanced. Make sure to balance the charges as well.

E. Incorrect. 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 Mg2+

Answer: (B) Mg + 2Na+ 2Na + Mg2+

Solution

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