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Chapter 8 Sample Test

Chapter 8 Sample Test. Warning This PowerPoint is not printer friendly. Reminder. Make sure you know the 5 types of reactions: Combination / synthesis Decomposition Single Replacement Double replacement Combustion.

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Chapter 8 Sample Test

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  1. Chapter 8 Sample Test

  2. WarningThis PowerPoint is not printer friendly.

  3. Reminder • Make sure you know the 5 types of reactions: • Combination / synthesis • Decomposition • Single Replacement • Double replacement • Combustion

  4. 1. Aluminum chloride and bubbles of hydrogen gas are produced when metallic aluminum is placed in hydrochloric acid. What is the balanced equation for this reaction?

  5. First you need to figure out what the problem is stating: • Remember: Reactants Products Aluminum chloride and bubbles of hydrogen gas are produced. Produced means products when metallic aluminum is placed in hydrochloric acid So these must be the reactants.

  6. So lets rewrite the equation: • aluminum is placed in hydrochloric acid • Aluminum chloride and bubbles of hydrogen gas are produced Al(s) + HCl(aq) As a chemical equation AlCl3 + H2

  7. Time to balance Al(s) + HCl(aq) 2 2 2 AlCl3 + H2 3 6

  8. Final Answer 2Al(s) + 6HCl(aq) 2AlCl3 + 3H2

  9. 2. When the equation Fe + Cl2 → FeCl3, is balanced, what is the coefficient for Cl2? Fe + Cl2 → FeCl3 2 2 3 Answer = 3

  10. 3.When the following equation is balanced, what is the coefficient for HCl? Mg(s) + HCl(aq) → MgCl2(aq) + H2↑ 2 And Yes, Sometimes it is that easy. Answer = 2

  11. 4. When the following equation is balanced, KClO3(s) KCl(s) + O2(g), the coefficient of KClO3 is ___. KClO3(s) KCl(s) + O2(g) 2 2 3 Answer = 2

  12. 5. Write a balanced equation to represent the decomposition of lead(IV) oxide. First you need to know that lead(IV) oxide Has the formula of PbO2. Next you need to remember that decomposition Means, breaking down into pieces. And look It’s already Balanced! So the PbO2 is decomposing into lead and oxygen. PbO2 Pb + O2

  13. 6. What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodide? First you need to remember that bromine and Iodine are diatomic gases. Br2 I2 2nd you need to know sodium iodide is NaI. This is a single displacement reaction. NaBr + I2 Br2 + NaI

  14. NaBr + I2 2 2 Br2 + NaI We have 2 Bromines on the left hand side of this equation, so we need to place a 2 in front of the NaBr to balance the equation. By placing the 2 in front of the NaBr, the number of sodiums was doubled, so we need to place a 2 in front of the NaI on the left side to balance the equation.

  15. 2 2 7. Balance the following equation. NaClO3 → NaCl + O2↑ The sodium and Chlorine are balanced. There are 3 oxygens on the left and only 2 on the right. We need to multiply the odd number of oxygens by an even number. By multiplying the right side by 2 we double the number of sodium and Chlorines. In order to balance this change, we need to place a 2 in front of the NaCl on the right hand side of the equation.

  16. 2 2 3 NaClO3 → NaCl + O2↑ Now we need to balance the oxygen. We have a total of 6 on the left hand side. To balance the equation we need to mutiply the “O2”s on the right hand side by 3. And now it is balanced!

  17. 2 8. Balance the following equation. Mg + H3PO4→ Mg3(PO4)2↓ + H2↑ The first thing we need to notice is that there are 3 hydrogens on the left and only 2 on the right. We know that an even # can’t divide perfectly into an odd number, so we need to multiply the H3PO4 by 2 so the left side will have an even # of hydrogens.

  18. 2 3 3 8. Balance the following equation. Mg + H3PO4→ Mg3(PO4)2↓ + H2↑ Now we have 6 hydrogens on the left. To balance this, we need to multiply the “H2” on the right by 3. Finally we need to multiply the Mg on the Left by 3. This will balance the magnesium

  19. 2 2 2 9. Balance the following equation. (NH4)2CO3 + NaOH → Na2CO3 + NH3 + H2O Notice that the hydrogens are spead out all Over the place. Wait and balance them last. By placing this two here our nitrogens become balanced. And now everything is balanced. luckily for us we never had to worry about the balancing the hydrogen. The carbonate polyatomic ions appear to be balanced, but we need to place a 2 here to balance the sodium. By multiplying the NaOH by 2 we added another Oxygen atom to the left hand side of this equation To balance this out we can multiply the H2O on The right by 2.

  20. ∆ 3 3 3 10. Balance the following equation. C3H6 + O2 → CO(↑) + H2O(↑) There are 3 carbons on the left so we need To place a 3 in front of the CO We have 6 Hydrogens on the left, so We need to place a 3 in front of the water. Now we have 6 oxygens on the right, so We need to multiply the O2 by 3. And now it is balanced.

  21. 2 11. Balance the following equation. Ba + H2O → Ba(OH)2 + H2(↑) The barium is already balanced. However there are only 2 hydrogens on The left, but 4 on the right. To fix this we need to place a 2 in front of The water. And then everything balances!

  22. 2 4 3 12. Balance the following equation. Au2O3→ Au + O2(↑) Finally, we need to multiply the O2 on the Right by 3. In order to balance the gold. We need to multiply the gold on the Right side of the equation by 4. First we need notice that there is an odd number of oxygens on the left and an even number on the right. In order to balance the equation we will need to put a 2 in front of the gold oxide. And now it is balanced!

  23. 3 3 2 13. Balance the following equation. Na3PO4 + ZnSO4→ Na2SO4 + Zn3(PO4)2↓ First notice that there is an odd number of sodiums on the left and an even number on the right. In order to make both numbers even we should place a 2 in front of the sodium phosphate. Now we have 3 sulfates (SO4)s on the right so we need to place a 3 in front of the zinc sulfate to balance the sulfates. Now we need to place a 3 in front of the Sodium sulfate to balance the sodiums. Notice SO4s and PO4s behave like atoms And now it is balanced!

  24. 6 8 2 4 3 7 14. Balance the following equation. C3H8 + O2→ CO↑ + H2O↑ Now the equation is balanced. Lets start by placing a 3 here to balance the Carbons. Now we need to place a 4 here to balance The hydrogens Finally, we place a 2 here to balance out the carbons and hydrogens. To balance, the oxygens we need to multiply The “O2” on the left by 7 Now we need to balance our oxygens. We have 7 on the right and 2 on the left. If we double our products our oxygens on The right hand side will become 14.

  25. 15. Write a balanced net ionic equation for the following reaction. H3PO4(aq) + Ca(OH)2(aq) Ca3(PO4)2(aq) + H2O(I) 3 2 6 Notice the Hydrogens are all over the place. Balance those last. There is 1 phosphate on the left, and 2 on the right. So we will place a 2 here to balance them. Now lets balance the calciums by placing a 3 here. Finally we can balance the oxygens and Hydrogens by placing a 6 here. Now the equation is balanced!

  26. 15. Now we need to cancel out the spectator ions. H3PO4(aq) + Ca(OH)2(aq) Ca3(PO4)2(aq) + H2O(I) 3 2 6 2 H3 + (OH)2(aq) H2O(I) 3 6 Final Net ionic equation

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