Fundamentals%20L%26M%20Solns

Fundamentals%20L%26M%20Solns - L.1 2 Na 2 S2 O3(aq AgBr(s...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: L.1 2 Na 2 S2 O3 (aq) + AgBr(s) NaBr(aq) + Na 3 [Ag(S2 O3 ) 2 ](aq) (a)moles of Na 2 S2 O3 needed to dissolve 1.0 mg AgBr 1 g AgBr 1 mol AgBr 2 mol Na 2 S2 O3 = 1.0 mg AgBr 1000 mg AgBr 187.78 g AgBr 1 mol AgBr = 1.1 10-5 mol Na 2 S2 O3 (b)mass of AgBr to produce 0.033 mol Na 3 [Ag(S2 O3 ) 2 ] 187.78 g AgBr 1 mol AgBr = 0.033 mol Na 3 [Ag(S2 O3 ) 2 ] 1 mol Na 3 [Ag(S2 O3 ) 2 ] 1 mol AgBr = 6.2 g AgBr L.5 2 C57 H110 O6 (s) + 163 O 2 (g) 114 CO 2 (g) + 110 H 2 O(l) 1 mol fat 110 mol H 2 O 18.02 g H 2 O (454 g fat) 891.44 g fat 2 mol fat 1 mol H 2 O (a) = 505 g H 2 O mol fat 163 mol O 2 32.00 g O 2 1 (454 g fat) 891.44 g 2 mol fat 1 mol O 2 (b) = 1.33 103 g O 2 (a) HCl + NaOH NaCl + H 2 O L.9 0.234 mol HCl 1 mol NaOH 17.40 mL = 0.00407 mol 1000 mL 1 mol HCl concentration of NaOH = 0.00407 mol = 0.271 mol L-1 -3 15.00 10 L L mol (b) (0.271 mol -1 )(0.01500 L) (40.00 g -1 ) = 0.163 g NaOH L.13 HX(aq) + NaOH(aq) NaX(aq) + H 2 O(l) 0.750 mol NaOH (68.8 mL) = 0.0516 mol NaOH 1000 mL NaOH 3.25 g HX corresponds to 0.0516 mol NaOH used 3.25 g = 63.0 g mol-1 = molar mass of acid 0.0516 mol L.21 (a) S2O32- is both oxidized and reduced. (b) Find the number of grams of thiosulfate ion in 10.1 mL of solution. g HSO- (aq) 55.0 g HSO- 1.45 2 - 3 3 ? g S2 O3 - = 10.1 mL HSO3 (aq) mL HSO- (aq) 100 g HSO- (aq) 1 3 3 - 2 2 mol HSO3 1 mol S2 O3 - 112.0 g S2 O3 - 1 - - 2 81.0 g HSO3 1 mol HSO3 1 mol S2 O3 - = 11.1 g S O 2- present initially 2 3 L.27 First equation must be balanced as following: Fe + Br2 FeBr2 3 FeBr2 + Br2 Fe3Br8 Fe3Br8 + 4 Na2CO3 8 NaBr + 4 CO2 + Fe3O4 The masses of Fe are needed to produce 2.5 t of NaBr: 1000 kg 1000 g 1 mol NaBr 1 mol Fe3Br8 2.50 t 1 t 1 kg 102.9 g NaBr 8 mol NaBr mol FeBr2 1 mol Fe 55.84 g Fe 3 5 = 5.09 10 g Fe 1 mol Fe3Br8 1 mol FeBr2 1 mol Fe = 509 kg Fe L.31 Total mass of tin oxide is: 28.35g 26.45 g = 1.90 g 1 mol Sn -2 = (a) 1.50 g Sn 1.264 10 mol Sn 118.71 g Sn mol O 1 = moles of O: (1.90 - 1.50)g 0.025 mol O 16.00 g O mole ratio of Sn:O is 1:2 empirical formula: SnO2 (b) tin(IV) oxide M.1 CaCO3 (s) 3 CaO(s) + CO 2 (g) theoretical yield: 1 mol CaCO3 1 mol CO 2 44.01 g CO 2 (42.73 g CaCO3 ) 100.09 g CaCO3 1 mol CaCO3 1 mol CO 2 = 18.79 g CO 2 actual yield: 17.5 g 3 100% = 93.1% yield 18.79 g M.5 (a) P4 (s) + 3 O 2 (g) P4 O6 (s) + 2 O 2 (g) P4 O6 (s) P4 O10 (s) In the first reaction, 5.77 g P4 uses 1 mol P4 3 mol O 2 32.00 g O 2 (5.77 g P4 ) 123.88 g P4 1 mol P4 1 mol O 2 excess O 2 = 5.77 g - 4.47 g O 2 = 1.30 g O 2 In the second reaction, 5.77 g P4 uses 5.77 g P4 1 mol P4 O6 2 mol O 2 32.00 g O 2 -1 123.88 g mol P4 1 mol P4 1 mol P4 O6 1 mol O 2 = 2.98 g O 2 limiting reagent: O 2 1.30 g O 2 1 mol P4 O10 283.88 g P4 O10 = (b) 5.77 g P4 O10 -1 32.00 g mol O 2 2 mol O2 1 mol P4 O10 = 4.47 g O 2 (g) 1.30 g O 2 1 mol P4 O6 219.88 g P4 O6 -1 32.00 g mol O 2 2 mol O 2 1 mol P4 O6 (c) = 4.47 g P4 O6 used In the first reaction, 5.77 g P4 produces 5.77 g P4 219.88 g P4 O6 1 mol P4 O6 -1 123.88 g mol 1 mol P4 O6 1 mol P4 excess reagent: 10.2 g - 4.47 g = 5.7 g P4 O6 M.9 1 mol CO 2 1 mol C (0.682 g CO 2 ) 44.01 g CO2 1 mol CO 2 = 0.0155 mol C = 10.2 g P4 O6 (0.0155 mol C) (12.01 g -1 C) = 0.186 g C mol 1 mol H 2 O 2 mol H (0.174 g H 2 O) = 0.0193 mol H 18.02 g H 2 O 1 mol H 2 O -1 (0.0193 mol H) (1.0079 g mol H) = 0.0195 g H 1 mol N 2 2 mol N (0.110 g N 2 ) = 0.007 85 mol N 28.02 g N 2 1 mol N 2 (0.007 85 mol N) (14.01 g -1 N) = 0.110 g N mol mass of O = 0.376 g - (0.186 g + 0.0193 g + 0.110 g) = 0.061 g O 0.061 g O = 0.0038 mol O 16.00 g O Dividing each amount by 0.0038 gives C : H : N : O ratios = 4.1 : 5.1 : 2.1 : 1. The empirical formula is C 4 H 5 N 2 O. The molecular mass of caffeine is 194 g mol -1 . Its empirical mass is 97.10 g mol-1 . molecular formula = 2 empirical formula = C8 H10 N 4 O 2 2 C8 H10 N 4 O 2 (s) + 19 O 2 (g) 16 CO 2 (g) + 10 H 2 O(l) + 4 N 2 (g) M.13 3 Ca(NO3 ) 2 (aq) + 2 H 3 PO 4 (aq) Ca 3 (PO 4 ) 2 (s) + 6 HNO3 (aq) (a)The solid is calcium phosphate, Ca 3 (PO 4 ) 2 . 1 mol Ca(NO3 ) 2 2 mol H 3 PO4 (206 g Ca(NO3 ) 2 ) 164.10 g Ca(NO3 ) 2 3 mol Ca(NO3 ) 2 (b) 97.99 g H 3 PO 4 = 82.01 g H 3 PO 4 1 mol H 3 PO 4 Therefore, Ca(NO3 ) 2 is the limiting reagent. 1 mol Ca(NO3 ) 2 1 mol Ca 3 (PO 4 ) 2 (206 g Ca(NO3 ) 2 ) 164.10 g Ca(NO3 ) 2 3 mol Ca(NO3 ) 2 310.18 g Ca 3 (PO 4 ) 2 = 130 g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 M.19 Determine the number of moles of each element present in the compound and then find their ratios to get the subscripts for the empirical formula. 0.055 g Cl = 1.55 10-3 mol Cl -1 35.453 g mol 0.0682 g CO 2 = 1.55 10-3 mol CO 2 = mol C -1 44.0 g mol 0.0140 g H 2 O = 7.78 10-4 mol H 2O = 1.56 10-3 mol H -1 18.01 g mol 0.100 g compound 12.0 g C 1.0079 g H - 0.055 g Cl +1.55 -3 mol 10 + 1.55 -3 mol 10 mol mol = 0.0247 g O 1 mol -3 = 1.55 10 mol O 16.00 g The mole ratio is 1:1:1:1, so the empirical formula is CHOCl. As 0.100 g of the compound contains 1.5510-3 moles of each element, its molar mass is 0.100 g = 64.52 g mol -1 . The molar mass of the 1.5510-3 mol empirical formula is 64.47 gmol-1, so the molecular formula is CHOCl. M.21 HA + XOH H2O + XA 2.45 g HA Moles of HA: = 0.0106 mol 231 mol-1 g 1.50 g XOH Moles of XOH: = 0.0120 mol 125 mol-1 g Molar mass of XA = (231-1.0) + (125-17) = 338 gmol-1 2.91 g XA -3 Mole of XA produced: = -1 8.61 10 mol 338 mol g Theoretical yield: (based on the limiting reagent, HA) 1 mol XA 0.0106 mol HA = 0.0106 mol XA 1 mole HA 8.61 10-3 mol XA Percentage yield: 100% = 81.2% 0.0106 mole XA ...
View Full Document

This note was uploaded on 04/24/2008 for the course CHEM 6A taught by Professor Pomeroy during the Fall '08 term at UCSD.

Ask a homework question - tutors are online