15.0 g Fe O 2 mol Fe 55.8 g mol Fe = g

Transcription

1 CHAPTER Practice Questions.1 1 Mg, O, H and Cl (on each side).. BaCl (aq) + Al (SO ) (aq) BaSO (s) + AlCl (aq) mol 106 g mol 1 = 1. g 15.0 g Fe O mol Fe 55.8 g mol Fe = 10.9 g g mol FeO 1 mol FeO 1.7 g Fe =.9 %N = 0.117/ = 5.9% %O = 0.05/ = 7.06% Since the sum of %N and %O equals 100%, there are no other elements present..11 It is convenient to assume that we have 100 g of the sample, so that the % mass represents masses. Thus there is. g of Na,.6 g of S and ( ) = 5.00 g of O. Convert these masses to moles:. g Na mol Na = = 1.0 mol 1.00 g mol Na.6 g S mol S = = 0.70 mol 1.06 g mol S 5.0 g O mol O = =.81 mol g mol O Next, we divide each of these mole amounts by the smallest in order to deduce the simplest whole number ratio: For Na: 1.0 mol/0.70 mol =.0 For S: 0.70 mol/0.70 mol = 1.0 For O:.81 mol/0.70 mol =.0 The empirical formula is therefore Na SO. 5 mol O mol O = mol CO = mol mol CO.1.15 First determine the mass of O that is required to react completely with the given mass of ammonia:

2 0.00 g NH 5 mol O 1 g O = 1.00 g mol O 17.0 g mol NH mol NH = 70.6 g Since this is more than the amount that is available, we conclude that oxygen is the limiting reactant, and therefore: 0.00 g O mol NO 1 g NO = g mol NO.00 g mol O 5 mol O = 0.01 g.17 mol Na SO =.550 g Na SO g mol NaSO = mol 1 L L solution = ml = L 1000 ml moles solute mol Na SO M = 0.98 M L solution L solution.19 (V dil )(M dil ) = (V conc )(M conc ) (100 ml)(0.15 M) = (V conc )(0.500 M) V conc = (100 ml)(0.15 M)/(0.500 M) = 5.0 ml Therefore, mix 5.0 ml of M H SO with water to make 100 ml of total solution..1 FeCl Fe + + Cl M Fe mol FeCl 1mol Fe = 1 L FeCl soln 1mol FeCl = 0.0 M Fe +. The balanced net ionic equation is: Fe + (aq) + OH (aq) Fe(OH) (s). First determine the number of moles of Fe + present: mol FeCl 1mol Fe Fe 1000 ml solution 1mol FeCl ml FeCl solution = mol Now, determine the amount of KOH needed to react with the Fe + :

3 mol 1mol + OH KOH 1000 ml solution 10 mol Fe = 60.0 ml KOH mol Fe 1mol OH mol KOH 0.76 g CaSO 1mol CaSO.5 (a) 16.1 g CaSO 1mol CaSO 1mol Ca + = mol Ca + Since all of the Ca + is precipitated as CaSO, there were originally moles of Ca + in the sample. Review Problems.1 A chemical equation is balanced when there is the same number of each kind of atom on both the reactant and product side of the equation. This condition arises from the law of conservation of matter.. Coefficients.5 (a) Student B is correct. Student A wrote a properly balanced equation. However, by changing the subscript for the product of the reaction from an implied one, NaCl, to NaCl, this student has changed the identity of the product. When balancing chemical equations, never change the value of the subscripts in chemical formulae..7 Mole not mass is the fundamental unit of chemistry. Stoichiometry is given by a balanced chemical equation, which is directly interpreted in moles. Mass (g).9 Amount of substance (moles) 1 Molar mass (g mol ).11 There are the same number of molecules in.5 moles of H O and.5 moles of H..1 The statement 1.0 mole of oxygen does not indicate whether this is atomic oxygen, O, or molecular oxygen, O. The statement 6 g of oxygen is not ambiguous because the source of oxygen is not important..15 When balancing a chemical equation, changing the subscripts of a chemical formula changes the identity of the corresponding compound..17 To determine the number of grams of sulfur that reacts with 1 gram of arsenic, the stoichiometric ratio of arsenic to sulfur in the compound is needed together with the molar masses of sulfur and arsenic..19 H O H O + O

4 .1 mmol 1 mol 1000 ml 1000 mol mol ml 1000 mmol 1 L 1000 L L. On dilution, the number of moles of HNO in the solution has not changed but the concentration has decreased since water has been added..5 Qualitative analysis is the use of experimental procedures to determine the elements that are present in a substance. Quantitative analysis determines the percentage composition of a compound or the percentage of a component in a mixture. Qualitative analysis answers the question, what is in the sample? Quantitative analysis answers the question, how much is in the sample?.7 The charge on Co is incorrect and the physical states of the reactants and products are not given. Balanced equation is: Co + (aq) + HPO (aq) Co (PO ) (s) + H + (aq)..9 (CH ) S + 90 S0 + CO + 6H O.1 CH OH + N O CO + H O + N. H NCHO + 5O CO + 6H O + N.5 (a) Ca(OH) + HCl CaCl + H O AgNO + CaCl Ca(NO ) + AgCl Fe O + C Fe + CO (d) NaHCO + H SO Na SO + H O + CO (e) C H O 8CO + 10H O.7 O is the limiting reagent mol Na atoms Na = mole Na atoms Na.1 Based on the balanced equation: NH (g) N (g) + H (g) Hence, the conversion factors are: 1mol N mol NH mol H and mol NH To determine the moles produced, convert starting moles to end moles : 1mol N 0.15 mol NH mol N mol NH The moles of hydrogen are:

5 0.15 mol NH mol H 0.18 mol H mol NH. The empirical formula is NaTcO..5 The empirical formula is C H 6 O..7 (a) 1 g Fe (1.5 mol Fe)(55.85 g mol Fe) 75. g 1 g O (.5 mol O)(16.00 g mol O) 9 g 1 g Ca (0.876 mol Ca)(0.08 g mol Ca) 5.1 g 17.7 g Ni.9 mol Ni = g mol Ni = 0.0 mol.51 (a) g Ca (PO ) 1.5 mol Ca (PO ) g mol Ca (PO ) 88 g g Fe(NO ) 0.65 mol Fe(NO ) 1.87 g mol Fe(NO ) g g C H mol C H g mol CH10.9 g (d) g (NH ) CO 1.5 mol (NH ) CO g mol (NH ) CO 1 19 g.5. kg fertiliser atoms N 5 mol O g O 16.0 g mol O 0.7 g atoms mol N mol N mol Zn mol Au(CN).57 (a) (0.11 mol Au(CN) 1 ) 65.9 g mol Zn mol Au mol Au(CN) (0.11 mol Au(CN) 1 ) g mol Au =.6 g Zn = g Zn

6 mol Au(CN) 1 mol Zn (CN) 1 = (0.11 mol Zn) 9.0 g mol Au(CN) = 55 g Au.59 mol O 1000 g H O 1 moles O 1.70 mol 1.01 g mol HO mole HO 1 1 kg O kg O 1.70 moles O.00 g mol O 0.70 kg 1000 g O.61 Since more than this minimum amount is available, FeCl is present in excess and AgNO is the limiting reactant g (a) g NaCl = 0.15 L soln 0.00 mol L NaCl58. g mol NaCl = 1.6 g g C H 1 O 6 = 0.50 L soln mol L C 1 6H1O g mol C6H1O6 = 16. g g H SO = 0.50 L soln mol L H 1 SO g mol HSO = 6.1 g.67 The 5.0 ml of HSO must be diluted to 00 ml..69 (a) Cr(NO ) Cr + + NO mol Cr 1 mol Cr(NO ) M Cr = 0.5 mol L Cr(NO ) = 0.5 M 1 mol NO M NO = 0.5 mol L Cr(NO ) = 0.50 M 1 mol Cr(NO ) CuSO Cu + + SO mol Cu 1 mol CuSO M Cu = 0.10 mol L CuSO = 0.10 M 1 1 mol SO M SO = 0.10 mol L CuSO = 0.10 M 1 mol CuSO

7 Na PO Na + + PO mol Na 1 mol NaPO M Na = 0.16 mol L Na PO = 0.8 M 1 1 mol PO M PO = 0.16 mol L NaPO = 0.16 M 1 mol Na PO (d) Al (SO ) Al + + SO mol Al 1 mol Al (SO ) M Al = mol L Al (SO ) = 0.15 M 1 mol SO M SO = mol L Al (SO ) = 0. M 1 mol Al (SO ) mol Ca M Ca = mol L Ca(OH) = M 1 mol Ca(OH) 1 mol OH M OH = mol L Ca(OH) = 0.1 M 1 mol Ca(OH).71 ml NiCl soln 1 1 mol NiCl 1000 ml soln = 0.00 L soln0.15 mol L NaCO 1 mol NaCO 0.5 mol NiCl = 1.0 ml g NiCO 1 mol NiCO 1 mol NiCl 1 = 0.01 L NiCl soln mol L NiCl g mol NiCO = 0.6 g.7 1. ml.75 (a) 0.9 g For nitrate: M For Na + : 0.5 M For phosphate: 0.07 M Additional Exercises

8 1 mol AgCl 1 mol NaCl 1. g AgCl 1 mol AgCl = 0.11 g The entire sample was NaCl. 1 g NaCl = (0.77 g AgCl) 58. g mol NaCl V = 50 ml.81 (a) mol L Na PO L 1. mol L 1.8 The empirical formula of the compound is determined from the information given however, its molar mass is needed to determine the molecular formula..85 (a) 1. g ions The overall charge on CuAl 6 (PO ) (OH) 8 is zero and since the charge on aluminium, phosphate and hydroxide is +, and 1, respectively, the charge on Cu is (a) C =.6% H =.18% N = 1.8% O = 1.0% P = 18.% (d) atoms g g.89 Since one molecule of vitamin B 1 contains one atom of Co, then 1 mole of vitamin B 1 contains 1 mole of Co. Since the molar mass of Co is 58.9 g mol 1,.% of the mass of vitamin B 1 is 58.9 g. Hence the molar mass of vitamin B 1 is 158 g mol Md 0.17 mol L molecules.95 (a) 5.8 g S molecules 1.88 mol

9 (d) Since the density of water is 1.00 g ml 1, there is 0.05 g of Al (SO ).18H O per ml of solution..97 Remaining SiO mol SiO g mol.7 g Remaining C mol C 1.01 g mol 6.8 g Remaining Cl mol Cl g mol.55 g