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1、A 1.0 g sample of an alkaline earth metal M reacts completely with 0.8092 g of chlorine gas to yield and ionic salt with the formula MCl2. In the process 9.46 kJ of heat is released. What is the molecular mass and identity of the metal M?M + Cl2 = MCl21 mol of M reacts with 1 mol Cl20.8092g/71g/mol
2、= 0.0114 mol1 g M/gAt Wt M = 0.0114 molM = 87.7 g/molA check of the periodic table reveals that M is?SrIf 9.46 kJ/mol of heat was released, how much heat would be released if 1 mol of SrCl2 was formed?-9.46 kJ/g *87.7 g/mol = -829.6 kJ/mol1Sodium nitrite, NaNO2, is frequently added to processed meat
3、s as a preservative. The amount of nitrite in a sample can be determined by reducing the nitrite to nitric oxide (NO) in acid with excess iodide which forms I3- and the titrating the I3- liberated with thiosulfate (S2O3-2) to form iodide ion and S4O6-2. When a nitrite containing sample of meat (2.93
4、5g) was analyzed, 18.77 mL of 0.15 M Na2S2O3 solution was needed for the analysis. What is the mass percentage of sodium nitrite in the meat sample?NaNO2 + I- = NO (g) + Na + I3-oxidation state of N in NaNO2? +3In NO?+2oxidation state of I-?-1In I3-1/3e- +NaNO2 + 2H+ = NO(g) + H2O + Na+ reduction 3I
5、- = I3- + 2e- oxidation2NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2 Na+22NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2Na+Titrating the I3- liberated with thiosulfate (S2O3-2) forms iodide ion and S4O6-2.oxidation state of I3-?-1/3In I- ?-1oxidation state of S in S2O3-2?+2In S4O6-2 ?+2.5 I3- +2 e- = 3I-2
6、S2O3-2 = S4O6-2 +2 e- 2S2O3-2 + I3- = S4O6-2 + 3I-32NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2Na+2S2O3-2 + I3- = S4O6-2 + 3I-When a nitrite containing sample of meat (2.935g) was analyzed, 18.77 mL of 0.15 M Na2S2O3 solution was needed for the analysis. What is the mass percentage of sodium nitrite
7、in the meat sample? How many mols of sodium thiosulfate was consumed?0.01877 L * 0.15 M = 0.0028 mol Na2S2O3How many mol of I3- must have been present?0.0014 mol I3-Therefore 0.0014 mol of I3- must have been produced in the first reactionHow many mol of NaNO2 are needed to produce 0.0014 mol I3-?0.0
8、028 mol NaNO2MW NaNO2 = 69 g/mol0.0028 mol NaNO2* 69 g/mol = 0.193g0.193/2.935*100 = 6.6 %4Photochromic glasses which darken on exposure to light, contain a small amount of silver. When irradiated with light, the silver ion is reduced to silver metal by capture of an electron from chlorine to produc
9、e a chlorine atom and the glass darkens. The chlorine atom formed is prevented from diffusing away by the glass. When the light is removed, the silver metal loses an electron and silver chloride is reformed. If 310 kJ/mol of energy is necessary for the reaction to proceed, what wavelength of light i
10、s necessary?310 kJ/mol/6.02*1023 atoms/mol = 51.5*10-23 = 51.5*10-20 J/atomE = h*; 51.5*10-20 J/atom= 6.63*10-34 Js* = 7.77*1014 s-13*108 m/s = *3*108 cm/s = *7.77*1014 s-1 = 3.86*10-7 m *109 nanometers/m = 386 nm5386 nm6Three atoms have the following electronic configurations:a)1s22s22p63s23p1b)1s2
11、2s22p63s23p5c)1s22s22p63s23p64s1d)Which of the three has the largest Ei1? e)Which of the three has the smallest Ei4?AlClKClCl7A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M2O) and nitride (M3N). The reaction product was dissolved in water according to Rx 1 and 2
12、and titrated with 0.1M HCl requiring 96.8 mL for complete neutralization. What is the metal and composition?Rx 1: M2O + H2O = MOHRx 2: M3N + H2O = MOH + NH3HCl + MOH + NH3 = MCl + H2O + NH4Cl What is the first thing to do?balance the equations!Rx 1: 4M + O2 = 2M2ORx 2: 6M+ N2 = 2M3N1/2M2O + 1/2H2O =
13、 MOH1/3M3N + H2O = MOH + 1/3NH3Rx 1:M + O2 = M2O Rx 2 M+ N2 = M3N M + 1/4O2 = 1/2M2OM + 1/6N2 = 1/3 M3NMOH + HCl = MCl + H2OMOH +1/3NH3 + 4/3HCl = MCl + 1/3NH4Cl +H2O8A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M2O) and nitride (M3N). The reaction product was di
14、ssolved in water and titrated with 0.1M HCl requiring 96.8 mL for complete neutralization according to the following two reactions. Rx (1)with oxygen: 1 mol of M reacts with 1 mol of HClRx (2)with nitrogen: 1 mol of M reacts with 4/3 mol of HClHow much HCl would be required if M is Li:only Rx 1 occu
15、rs: only Rx 2 occurs:0.053/6.9 =0.00764 mol LiAmount of HCl consumed to endpoint =0.0764*1= 0.00764 mol0.0764*4/3= 0.01019 molHow much HCl would be required if M is Na:only Rx 1 occurs: only Rx 2 occurs:0.053/23 =0.0023 mol Na0.0023 *1= 0.0023 mol 0.0023 *4/3= 0.0031 mol0.0968 L*0.1M; 0.00968mol9A 0
16、.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M2O) and nitride (M3N). The reaction product was dissolved in water and titrated with 0.1M HCl requiring 98.6 mL for complete neutralization according to the following two reactions. Rx (1)with oxygen: 1 mol of M reacts w
17、ith 1 mol of HClRx (2)with nitrogen: 1 mol of M reacts with 4/3 mol of HClAmount of HCl consumed to endpoint = 0.0986L*0.1; 0.00986 molHow much HCl would it take if Rx 1 and Rx 2 contributed equally?0.5*.00764+0.5*0.01019 = 0.00891 mollet a be the fraction reacting by Rx 1a*(0.00764) + (1-a)(0.01019) = 0.00986; a = 0. 148 or 14.8 % by Rx1compared to 0.00968mol10
