Solutions to the Extra Problems for Chapter 7

Transcription

1 Solutions to the Extra Problems for Chapter 7. a. The answer is moles. The Periodic Table says that Si has a mass of 8.09 amu. That also tells us: mole Si = 8.09 g Si 00.0 g Si mole Si 8.09 g Si moles Si Since we can have only four significant figures in the answer, we have to round Dropping the second 9 means rounding 59 to 60. We have to keep the zero, however, to have four significant figures. b. The answer is 8.34 moles. The Periodic Table says that H O has a mass of: Mass of H O =.0 amu amu = 8.0 amu Remember, this is like addition, since the number of atoms in a molecule is exact. Thus, we use the rules of addition for significant figures, which means we report our answer to the same decimal place as the least precise number in the problem. Since both the numbers have their last significant digit in the hundredths place, our answer must as well. That means: mole H O = 8.0 g H O 50.0 g H O mole HO 8.0 g H O 8.34 moles H O c. The answer is.775 moles. The Periodic Table says that C 6 H O 6 has a mass of: Mass of C 6 H O 6 = 6.0 amu +.0 amu amu = 80.8 amu Remember, we use the rules of addition to determine the significant figures. All the numbers have their last significant figure in the hundredths place, so the answer must as well. That means: mole C 6 H O 6 = 80.8 g C 6 H O g C H O mole C H O moles C O 6H g C6HO6

2 . a. The answer is 73 g. The Periodic Table says that S has a mass of 3.06 amu. That also tells us: mole S = 3.06 g S 5.4 moles S 3.06 g S mole S 73 g S b. The answer is 8.9 g. The Periodic Table says that KCl has a mass of: Mass of KCl = 39.0 amu amu = amu That means: mole KCl = g KCl 0. moles KCl g KCl mole KCl 8.9 g KCl c. The answer is 8.5 g. The Periodic Table says that C H 6 O has a mass of: Mass of C H 6 O =.0 amu amu amu = amu That means mole C H 6 O = g C H 6 O.76 moles C H6O g CH6O mole C H O g C H O 3. The answer is.40 5 atoms. We don t know any relationship between grams and atoms, but we do know a relationship between grams and moles. The Periodic Table says that magnesium (Mg) has a mass of 4.3 amu. That also tells us: mole Mg = 4.3 g Mg g Mg mole Mg 4.3 g Mg 0.57moles Mg Since we know that a mole is , we can say:

3 This allows us to convert to the number of atoms: mole Mg = atoms Mg moles Mg 6.00 atoms Mg atoms Mg mole Mg 4. The answer is 4.50 molecules. We don t know any relationship between grams and molecules, but we do know a relationship between grams and moles. Since moles are a way of counting molecules, we should start there. The Periodic Table says that AlCl 3 has a mass of: Mass of AlCl 3 = 6.98 amu amu = amu Remember, we use the rules of addition to determine the significant figures. All the numbers have their last significant figure in the hundredths place, so the answer must as well. This tells us: mole AlCl 3 = g AlCl g AlCl mole AlCl moles AlCl g AlCl3 Since we know that a mole is , we can say: mole AlCl 3 = molecules AlCl 3 This allows us to convert to the number of molecules: moles AlCl 6.00 molecules AlCl molecules AlCl 3 mole AlCl3 5. The answers are 83 moles and,400 g. Since we know that a mole is , we can say: mole NH 3 = molecules NH 3 That allows us to convert between molecules and moles: molecules NH 3 mole NH molecules NH 3 83 moles NH 3 Now that we know moles, we can convert to grams, once we have the mass:

4 This tells us: Mass of NH 3 = 4.0 amu amu = 7.04 amu mole NH 3 = 7.04 g NH 3 83 moles NH 7.04 g NH,400 g N H 3 mole NH3 6. a. For the hydrated form, we write the chemical formula, a dot, and then the number of H O molecules per molecule of the anhydrous form. That means the hydrated form is Na CO 3 0H O. b. The answer is 85. g. Remember, each Na CO 3 can have ten water molecules. So, we need to know how many Na CO 3 molecules we have. To count molecules, we must convert from grams to moles: That means Mass of Na CO 3 =.99 amu +.0 amu amu = amu mole Na CO 3 = g Na CO g Na CO mole Na CO moles Na CO g NaCO3 We know that each Na CO 3 can have ten water molecules, so the moles of water molecules are ten times that: Moles of water = 0moles of Na CO 3 = moles = 4.7 moles Remember, you can t have a fraction of a molecule, so the 0 is exact. That means the only thing that matters in terms of significant figures is the 0.47, which has three, so our answer must have three. That s the number of moles of water that can be absorbed, but the question asked for grams, so we have to convert: This tells us Mass of H O =.0 amu amu = 8.0 amu mole H O = 8.0 g H O 4.7 moles H O 8.0 g HO mole H O 85. g H O

5 7. a. You need 0.0 moles. The chemical reaction gives us the relationship between the reactants and products in terms of moles: mole Na SO 4 = moles Na We can use that to convert between moles of Na SO 4 and moles of Na: 0.0 moles Na SO moles Na mole Na SO moles Na b. You need 0.0 moles. The chemical reaction gives us the relationship between the reactants and products in terms of moles: mole Na SO 4 = mole S We can use that to convert between moles of Na SO 4 and moles of S: 0.0 moles Na SO mole S mole Na SO moles S c. You need 0.0 moles. The chemical reaction gives us the relationship between the reactants and products in terms of moles: mole Na SO 4 = moles O We can use that to convert between moles of Na SO 4 and moles of O : 0.0 moles Na SO moles O moles O mole NaSO4 8. The answer is 7.5 moles. In this reaction, we know that HCl is the limiting reactant, because the other reactant is in excess. Thus, HCl determines the amount of products. The chemical equation tells us: moles HCl = mole H We can use this to convert between moles of HCl and moles of H : 5 moles HCl mole H moles HCl 7.5 moles H 9. The answer is 3,66 g. In this reaction, we know that we want to make 00.0 moles of water. The equation gives us a relationship between moles of water and moles of H 3 PO 4 : 3 moles H O = mole H 3 PO 4

6 We can use this to convert between moles of water and moles of H 3 PO 4 : 00.0 moles H O moles H PO moles H PO moles HO Unfortunately, we aren t done, because the mean author wants the answer in grams. Since we know the number of moles, we just have to convert to grams: This tells us: Mass of H 3 PO 4 = 3.0 amu amu amu = amu mole H 3 PO 4 = g H 3 PO moles H PO g H PO ,66 g H PO 3 4 mole H3PO4 0. The answer is g. We know that aluminum is the limiting reactant, because the other one is in excess. To use the chemical equation, however, we must get the amount of aluminum in moles. The Periodic Table tells us aluminum has a mass of 6.98 amu, which also tells us: Now we can convert to moles: mole Al = 6.98 g Al The equation tells us: 00.0 g Al mole Al moles Al 6.98 g Al moles Al = 3 moles Cu This allows us to convert to moles of Cu: moles Al 3 moles Cu moles Cu moles Al Unfortunately, we aren t done, because the problem asks for the amount of Cu in grams. Thus, we have to convert. The Periodic Table tells us Cu has a mass of amu, which means: mole Cu = g Cu moles Cu g Cu mole Cu g Cu

7 . The answer is 9.46 g. In this reaction, we don t need to know the limiting reactant, because we are trying to find out how much of one reactant is needed. To do that, however, we must get the amount of water in moles: That means Mass of H O =.0 amu amu = 8.0 amu mole H O = 8.0 g H O 5.0 g H O mole HO 8.0 g H O Now we can use the chemical equation to convert to NH 3 : 0.83 moles H O 0.83 moles H O 6 moles H O = 4 moles NH 3 4 moles NH moles NH 3 6 moles HO The problem wants to know the number of grams of NH 3, so there is still one conversion left: This tells us: Mass of NH 3 = 4.0 amu amu = 7.04 amu mole NH 3 = 7.04 g NH moles NH 7.04 g NH 9.46 g NH 3 mole NH3