# Stoichiometry 99 in Visual C#.NET Decode DataMatrix in Visual C#.NET Stoichiometry 99

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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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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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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