Chemistry Chapter 6 Practice Problems. To avoid confusion, there is no Iodine on this set. If you see Cl, it is Chlorine!

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1 Chemistry Chapter 6 Practice Problems The more you do, the better you ll be prepared Some contain some little tricks to keep you thinking, but generally they look very much like the test will. The test DEFINITELY won t be stacked with the hardest questions, but if you feel ok with the tougher ones here, you should have no problem with the test. To avoid confusion, there is no Iodine on this set. If you see Cl, it is Chlorine! My molar masses may be slightly off because I used a different periodic table. If this is the case, do not worry about it This is not a comprehensive review for this chapter, but should help quite a bit Basic Stoichiometry 1. For the balanced reaction Na + 2 S NaS 2 If I am given 5 moles of Na and plenty of S, how many moles of NaS 2 can I make? 2. For the balanced reaction 6 Mg + P 4 2 Mg 3 P 2 If I produce 4.5 moles of Mg 3 P 2, how many moles of Mg and P 4 did I use? 3. For the balanced reaction Ca(OH) HCl CaCl H 2 O If I have 92 g HCl and excess Ca(OH) 2, what mass of each product will I produce? 4. For the balanced reaction 12 HClO 4 + P 4 O 10 4 H 3 PO Cl 2 O 7 If I have 450 g P 4 O 10, what mass of each product will I produce? 5. For the balanced reaction 2 KMnO HCl 2 KCl + 2 MnCl H 2 O + 5 Cl 2 If I want g of MnCl 2, what mass of each reactant will I need? 6. Using the unbalanced equation Au 2 S 3 + H 2 Au + H 2 S If I use 224 g Au 2 S 3, how many grams of Au will I end up with? 7. Another common household chemical is bleach. It is also called Hypochlorite. It has the formula NaOCl. If I make 88 g of bleach through a simple formation reaction, what mass of each reactant is used?

2 8. If I am running a complete combustion reaction of 4.5 kg of C 4 H 10 in an open room, what mass of each product will I obtain? Limits 9. For the balanced reaction 4 K + O 2 2 K 2 O If I am given 11 moles of K and 6 moles of O 2, what is the limiting reactant. How much mass of K 2 O can I make? 10. For the balanced reaction 2 H 2 + O 2 2 H 2 O If I am given 8 g H 2 and 97 g O 2, what mass of H 2 O can I make? 11. Bleach (NaOCl) is formed by the balanced chemical reaction Cl NaOH NaCl + NaOCl + H 2 O If the Chlorox factory has 482 kg NaOH and 500 kg Cl 2, what mass of bleach is produced? 12. For the balanced reaction If I have 137 g of H 3 PO 4 and 102 g HCl, what mass of each product will I produce, and how much leftover of the reactants will I have? 13. For the balanced reaction 3 CCl + Mg + 3 O 2 MgCl CO 2 If I am given g CCl, g Mg, and 96.0 g O2, what mass of each product will I produce, and how much leftover of the reactants will I have? 14. For the balanced reaction C 3 H O 2 3 CO H 2 O If I have 812 g C 3 H 8 and 318 g O 2, what mass of each product will I produce, and how much leftover of the reactants will I have? 15. For the unbalanced reaction C 4 H 10 + O 2 C + H 2 O If I have 11 g C 4 H 10 and g O 2, what mass of each product will I produce, and how much leftover of the reactants will I have? 16. Using the unbalanced equation: If I have g NH 3 and g O 2, what mass of each product will I produce, and how much leftover of the reactants will I have? 17. If I want to make NaOH in a formation reaction, and have 14.2 kg Na, 987 g Hydrogen gas, and 9 kg Oxygen gas, how much NaOH can I make, what mass of NaOH will I produce, and how much leftover of the reactants will I have?

3 Gay-Lussac 18. For the balanced reaction H 2 (g) + Cl 2 (g) 2 HCl (g) If I made 33 L of Hydrogen Chloride gas, what volume of each reactant were used? 19. For the balanced equation 2 N 2 (g) + 3 O 2 (g) 2 NO 3 (g) If I have 36 L of O 2 and plenty of Nitrogen, what volume of NO 3 will be produced? 20. The balanced equation: C 3 H O 2 3 CO H 2 O is typically all gas at ordinary temperatures. If I have 13 L of C 3 H 8 and plenty of Oxygen, how much volume of each product will I produce? 21. When bleach (NaOCl) is heated past its boiling point of 101 C, it decomposes. If I have 2 L of gaseous bleach, what volume of each product will be produced? 22. Using the equation: If I have 41 L of Hydrogen gas and 38 L of Oxygen gas, what volume of water will I procude and how much leftover of the reactants will I have? Empirical and Molecular Formulas 23. A chemical is broken down and found to have: g Sn 64.2 g S g Te Find its empirical formula. 24. A chemical is formed from: g Ta 55.7 g B 82.6 g C g At Find the chemical s empirical formula. 25. The empirical formula of a compound is found to be VErY 2 NiCe 3. The molecular mass is found to be 2625 g/mole. What is the molecular formula?

4 Answers: 1. We actually start with a stoichiometry problem that s actually a bit easier than the ones we ve been doing in class. Usually we have to cross the mole bridge going grams1 --> moles1 --> moles2 --> grams2. But if it s in moles, and wants moles for an answer, we just have to do the middle step! The plenty of S just tells us we don t need to worry about limiting reactants. The units cancel correctly, the answer makes sense, and it has the proper # of sig figs (1). So 5 moles Na is my answer. There will be a few of these on the test to get you what will hopefully be some easy points 2. Basically the same as #1, just have some bigger coefficients and unround numbers to deal with. It doesn t matter at all whether we are given the reactants or the products, as long as there are no limits, we just jump into stoichiometry. Do it once for each answer requested This one does require sig fig cleanup. There are 2 sig figs in my given number, so I need 2 in each answer. So the answers are 14 moles Mg and 2.3 moles P 4. (remember stoichiometric coefficients are perfect can t have 1.9 atoms, so they don t effect sig figs only the number I give you and numbers you pull off the periodic table matter) 3. Ok, we move on to the true stoichiometry problems you ve gotten more used to hopefully We are given a mass we must turn it into moles before we can compare to other chemicals. From the periodic table, HCl = = g per mole. So Remember to keep all your digits until we get our final answers, then round. Now we convert to each product by relating the moles. And convert each back into a mass in grams. The molar mass of CaCl is * g = 1 mole. The molar mass of H 2O is g = 1 mole. Then just line them up as we ve been doing all year. Then round to sig figs. Starting number had 2, the molar masses we used all had more (always try to use decent sig figs from the periodic table if you can, since you don t want to needlessly limit your answer). So we round to 2: 140 g CaCl 2 and 45 g H 2O. Considering we had similar amounts of each atom, and Ca and Cl are both heavier than H and O, it makes sense most of our mass is found there. The number compares similarly to HCl (and we cannot check with conservation of mass because we don t know how much Ca(OH) 2 we had).

5 4. Same steps as the last problem. So I ll start going quicker. If you absolutely cannot figure out where I came up with a number, call or ask. I do make mistakes! Sig figs were 2 due to 2 in the original. Seems a bit more than the starting number but it makes sense actually, because there s gotta be a lot of HClO 4 because it has a coefficient of 12! So there we go 5. And one more standard problem 6. This one needs to be balanced first (always check to see if I told you it was balanced, or if I didn t, that it is) Au 2S H 2 2 Au + 3 H 2S

6 7. Here you must write the reaction (and balance) first 2 Na + O 2 + Cl 2 2 NaOCl And now we CAN check conservation of mass if we wish since we now have all values = 88. The reactants are the same as the product. Now you can be % sure you got it right! 8. And the hardest one. Because you have to write the balanced reaction for a combustion 2 C 4H O 2 8 CO H 2O The second minor changeup was that I used kg instead of grams. Remember 1 kg = 1000 g. Probably best to work in grams (we need the units to cancel once we get the molar mass as you cannot just use kg as units. It s g -> mole when you pull from the table. 9. Again, start with the softball. Find the limiting reactant by finding how many batches each would make using the coefficients. Usually you d have to convert to grams first, but I made it nice The limiting reactant is K since it makes the least batches. Always then use the limiting reactant to figure out the products.

7 10. And now the normal problems. Gotta compare to find the limiting reactant, as with any time you are given two reactant amounts H 2 is the limiting reactant. So now we just figure out how much product is made. Note that I m going to start combining the chains from this point forward (I forgot until now!!!). You can still do it part by part if you wish, though this may be cleaner than trying to keep numbers. But you have to be careful of the calculations 11. Same as the previous problem, find the limiting reactant, then use that to find the mass of the product of interest. It s in kg, which won t cancel, so we ll add that conversion factor to be in grams so it works. So we ll run out of NaOH first. Now we just use that to figure out our answer. 12. Just a bit more to do here. But it starts the same way HCl is the limiting reactant. Now we need to figure out each product and we ll also need to compare with H 3 PO 4 so we can find out how much is leftover The first two are the products produced. To figure out how much H 3 PO 4 is leftover, just subtract the user (which we just calculated) from the initial. 82 g H 3 PO 4 [line up, chop off!]. May as well use mass conservation to check = 157 g reactants used = 157 g products Good

8 13. Same old story. 3 parts doesn t change anything. It s a tie. Since it s a tie, they are both the limiting reactants, and we can use either one for our conversions Leftover = = 154 g MgCl 3. Conservation of mass? Reactant: = 262 Product: = 263. Minor error due to rounding (if you compare the original numbers, both are very near total). Is good. 14. Same as the prior problem again So C 3 H 8, is the limiting reactant. Leftover C 3 H 8 is 812 g 87.7 g = 724 g = , validates within margin of error. 15. Balance the reaction first. Then same old, same old. 2 C 4 H O 2 8 C + 10 H 2 O The limiting reactant in a close race is O 2. Leftovers C 4 H 10 = = 1 Conservation check?: = = good

9 16. Balance and go 4 NH O 2 4 NO + 6 H 2 O Yet again, it is O 2. Man I need to mix things up more! Leftovers NO 3 = = 72.7 g Conservation check?: = = good 17. Write the reaction first (formation). 2 Na + O 2 + H 2 2 NaOH Gosh, I just can t find enough Oxygen in these problems Leftovers Na = = 4.2 kg. H2 = hmm, if we subtract 1000, we got issues. We should truly be keeping our full numbers until completion for each part then chop it at the end. So = 419 = 400. Conversation of mass here is going to probably be pretty loose, because the numbers are so rounded 18. Easy Gas-Lussac stoichiometric coefficients, all gases, can directly relate as if they were moles. 19. Same as the previous one, easy. They d conserve except the rounding budges it off just a little.

10 Must write the reaction to do this problem. 2 NaOCl 2 Na + O 2 + Cl 2 Then just like the others! Other answers to come may not be exactly on test