Transcription of CHAPTER 3: ANSWERS TO ASSIGNED PROBLEMS
1 1 CHAPTER 3: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 9/17/08 Balance the following equations: (d) Al4C3 (s) + 12 H2O (l) 4 Al(OH)3 (s) + 3 CH4 (g) (e) 2 C5H10O2 (l) + 13 O2 (g) 10 CO2 (g) + 10 H2O (g) do C, H, then half fraction the O2; double everything to remove fraction (f) 2 Fe(OH)3 (s) + 3 H2SO4 (aq) Fe2(SO4)3 (aq) + 6 H2O (l) Write balanced chemical equations to correspond to each of the following descriptions: (b) When solid potassium chlorate is heated, it decomposes to form solid potassium chloride and oxygen gas. 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) (c) Solid zinc metal reacts with sulfuric acid to form hydrogen gas and an aqueous solution of zinc sulfate.
2 Zn (s) + H2SO4 (aq) H2 (g) + ZnSO4 (aq) Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) 2 Al (s) + 3 Cl2 (g) 2 AlCl3 (s) COMBINATION (b) C2H4 (g) + 3 O2 (g) 2 CO2 (g) + 2 H2O (g) COMBUSTION (c) 6 Li (s) + N2 (g) 2 Li3N (s) COMBINATION (d) PbCO3 (s) PbO (s) + CO2 (g) DECOMPOSITION (e) C7H8O2 (l) + 8 O2 (g) 7 CO2 (g) + 4 H2O (g) COMBUSTION Calculate the percentage by mass of oxygen in the following compound: (a) morphine, C17H19NO3 NOTE: USE 2 DP ATOMIC WTS.
3 AS DISCUSSED C 17 X = (X 100%) = C H 19 X = (X 100%) = H N 1 X = (X 100%) = N O 3 X = (X 100%) = O answer sum = sum = PERCENTS OF THIS TYPE ARE TYPICALLY REPORTED TO A TENTH. 2 Calculate the following quantities: (a) mass, in grams, of moles sucrose (C12H22O11) C 12 X = H 22 X = O 11 X = sum = g / mol is formula weight mol sucrose ( g / mol sucrose) = = g (3 SF) (b) moles of Zn(NO3)2 in g of this substance Zn 1 X = N 2 X = O 6 X = sum = g / mol g zinc sulfate ( 1 mol zinc sulfate / g) = = mol (5 SF) Determine the empirical formula of the compound with the following composition by mass.
4 Percent to mass Mass to mole Divide by small Multiply till whole (c) Na, Al, and F g Na ( 1 mol Na / g) = mol Na / = = 3 g Al ( 1 mol Al / g) = mol Al / = = 1 g F ( 1 mol F / g) = mol F / = = 6 EMPIRICAL FORMULA IS Na3 AlF6 Determine the empirical formula of the compound with the following composition by mass: (c) C, H, % N, O g C ( 1 mol C / g )= / = X 2 = = 12 g H ( 1 mol H / g ) = / = X 2 = = 12 g N ( 1 mol N / g) = / = X 2 = = 2 g O ( 1 mol O / g) = / = X 2 = = 3 EMPIRICAL FORMULA IS C12H12N2O3 3 Determine the empirical and molecular formula for the following substance: (a) Styrene, a compound substance used to make Styrofoam cups and insulation, contains C and H by mass and has a molar mass of 104 g/ mol.
5 G C ( 1 mol C / g )= / = = 1 g H ( 1 mol H / g ) = / = = 1 EMPIRICAL FORMULA IS "CH" Since CH has a formula weight of , divide molar mass by that: 104 g/mol / g/mol = or 8; 8 will serve as a multiplier MOLECULAR FORMULA IS C8H8 Hydrofluoric acid, HF(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the HF(aq). Sodium silicate (Na2 SiO3), for example, reacts as follows: Na2 SiO3 (s) + 8 HF (aq) H2 SiF6 (aq) + 2 NaF (aq) + 3 H2O (l) (a) How many moles of HF are needed to react with mol of Na2 SiO3?
6 Mol Na2 SiO3 ( 8 mol HF / 1 mol Na2 SiO3 ) = mol HF (3 SF) (b) How many grams of NaF form when mol of HF reacts with excess Na2 SiO3? mol HF ( 2 mol NaF / 8 mol HF) ( NaF / 1 mol NaF) = = g (3 SF) (c) How many grams of Na2 SiO3 can react with g of HF? g HF ( 1 mol HF / g HF) ( 1 mol Na2 SiO3 / 8 mol HF) ( g / 1 mol Na2 SiO3 ) = = g Na2 SiO3 (3 SF) 4 Sodium hydroxide reacts with carbon dioxide as follows: 2 NaOH (s) + CO2 (g) Na2CO3 (s) + H2O (l) Which reagent is the limiting reactant when mol NaOH and mol CO2 are allowed to react?
7 Choose a product of interest, run both scenarios. NaOH scenario mol NaOH ( 1 mol Na2CO3 / 2 mol NaOH) = mol Na2CO3 CO2 scenario mol CO2 ( 1 mol Na2CO3 / 1 mol CO2 ) = mol Na2CO3 Since the NaOH scenario gives less product, the NaOH is the limiting reactant. How many moles of Na2CO3 can be produced? mol of Na2CO3 can be produced. The other scenario will not happen. When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained: C6H6 + Br2 C6H5Br + HBr (a) What is the theoretical yield of bromobenzene in this reaction when g of benzene reacts with g of bromine?
8 Do not know the LR. Run both scenarios. Beware of diatomic bromine! C6H6 scenario ( g C6H6 ) ( 1 mol C6H6 / C6H6) ( 1 mol C6H5Br / 1 mol C6H6) ( g C6H5Br / 1 mol C6H5Br) = = g C6H5Br (3 SF) Br2 scenario g Br2 ( 1 mol Br2 / g Br2 ) ( 1 mol C6H5Br / 1 mol Br2) ( g C6H5Br / 1 mol C6H5Br) = g = g C6H5Br (3 SF) Since the benzene scenario gave less product, this C6H6 is the Limiting Reactant. The g represents the theoretical yield for this reaction. (b) If the actual yield of bromobenzene was g, what was the percentage yield? % yield = (actual / theoretical) X 100% ( g actual / g theory) X 100% =
