Stoichiometry 99 in Visual C#.NET

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Stoichiometry 99
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12 E The balanced chemical equation is: 2 Au2O3 4 Au + 3 O2 (221 g Au2O3)(l mol Au2O3/442 g Au2O3)(3 mol O2/2 mol Au2O3) = 0750 mol O2 13 D The balanced equation is: 4 C4H11N(l) + 27 O2(g) 16 CO2(g) + 22 H2O(1) + 2 N2(g) 14 E The KMnO4 is the limiting reagent Each mole of KMnO4 will produce a mole of MnSO4 15 D The balanced equation is: 2 KClO3 2 KCl + 3 O2 (40 mol KClO3)(3 mol O2/2 mol KCIO3) = 60 mol O2 16 E The balanced equation is: 2 C8H18 + 25 O2 16 CO2 + 18 H2O (100 mol C8H18)(25 mol O2/2 mol C8H18) = 125 mol O2 17 E (0200 mol Ca)(l mol H2/l mol Ca)(224 L at STP/l mol H2)= 448 L
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3 2 2 18 C There are 500 10 mol of CrO4 and an equal number of mol of SnO2 Thus 2 SnO2 is the limiting reactant (larger coefficient in the balanced reaction) 3 3 2 2 (500 10 mol SnO2 )(2 mol OH /3 mol SnO2 ) = 333 10 mol OH
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19 D The volume of water is irrelevant 020 mol of KBr will require 010 mol of Pb(NO3)2 020 mol of MgBr2 will require 020 mol of Pb(NO3)2 Total the two yields
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U Free-Response Questions
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Answer the following questions You have 15 minutes, and you may use a calculator and the tables in the back of the book A sample of a monoprotic acid was analyzed The sample contained 400% C and 671% H The remainder of the sample was oxygen a Determine the empirical formula of the acid b A 02720-g sample of the acid was titrated with standard NaOH solution Determine the molecular weight of the acid if the sample required 4500 mL of 01000 M NaOH for the titration c A second sample was placed in a flask The flask was placed in a hot water bath until the sample vaporized It was found that 118 g of vapor occupied 3000 mL at 100 C and 100 atmospheres Determine the molecular weight of the acid d Using your answer from part a, determine the molecular formula for part b and for part c e Account for any differences in the molecular formulas determined in part d
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100 U STEP 4 Review the Knowledge You Need to Score High
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a The percent oxygen (533%) is determined by subtracting the carbon and the hydrogen from 100% For C: 400/1201 = 333 For H: 671/1008 = 666 For O: 533/1600 = 333 Divide each of these by the smallest (333) C=l H=2 O=1
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This gives the empirical formula: CH2O You get 1 point for correctly determining any of the elements, and 1 point for getting the complete empirical formula correct b Using HA to represent the monoprotic acid, the balanced equation for the titration reaction is: HA + NaOH NaA + H2O The moles of acid may then be calculated: (4500 mL NaOH)(01000 mol NaOH/1000 mL) (1 mol HA/1 mol NaOH) = 3 4500 10 mol HA The molecular mass is:
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3 02720 g/4500 10 mol = 6044 g/mol
You get 1 point for the correct number of moles of HA (or NaOH) and 1 point for the correct final answer c This may be done in several ways One way is to use the ideal gas equation This will be done here The equation and the value of R are given in the exam booklet First find the moles: n = PV /RT Do not forget, you MUST change to kelvin
n = (100 atm)(3000 mL)(l L/1000 mL)/(00821 L atm/mol K)(373 K) 3 n = 980 10 mol
3 The molecular mass is: 118 g/980 10 mol = 120 g/mol
You get 1 point for getting any part of the calculation correct and 1 point for getting the correct final answer d The approximate formula mass from the empirical (CH2O) formula is: 12 + 2(1) + 16 = 30 g/mol For part b: (6044 g/mol)/(30 g/mol) = 2 Molecular formula = 2 Empirical Formula = C2H4O2 For part c: (120 g/mol)/(30 g/mol) = 4 Molecular formula = 4 Empirical Formula = C4H8O4 You get 1 point for each correct molecular formula If you got the wrong answer in part a, you can still get credit for one or both of the molecular formulas if you used the part a value correctly
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