Exam I Sample Exam Solutions

Exam I Sample Exam Solutions - Department of Chemistry CHM...

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Department of Chemistry CHM 1220/1225 Exam I – Sample Exam – Solutions Note: The exam may not resemble this sample; it will help you prepareto complete the sample exam. 1. We are given 3 significant figures, so we need to used four digits in the molar mass. Our answer must have three significant figures. a. BaCl 2 (aq) + H 2 SO 4 (aq) Æ BaSO 4 (s) + 2 HCl(aq) 2 2 2 2 2 2 3 2 2 2 2 4 2 2 4 2 4 2 4 2 3 37 9 1 3 208 10 50 4 80 1 10 0 25 10 50 4 10 50 4 1 1 50 4 10 0 10 BaCl g . BaCl mol BaCl g . BaCl mol x . . c BaCl M . L x . BaCl mol x . BaCl mol x . SO H mol BaCl mol SO H L SO H mol . SO H L x . . b = = = 2. Cr 2 O 3 + 3 CCl 4 –––> 2 CrCl 3 + 3 COCl 2 Molar masses (g/mol): 152.0 153.8 98.92 We were given 3 significant figures, so we use molar masses with 4 significant figures. a. The calculations below give the mass of product that would be obtained if all that reactant is used up. 2 2 2 3 2 4 4 4 4 2 2 2 3 2 2 3 2 3 2 3 2 1 16 1 92 98 3 3 8 153 1 0 25 5 19 1 92 98 1 3 0 152 1 0 10 COCl g . COCl mol COCl g . O Cr mol CCl mol CCl . CCl mol CCl g . COCl g . COCl mol COCl g . O Cr mol COCl mol O Cr . O Cr mol O Cr g . = = CCl 4 is the limiting reactant; 16.1 g COCl 2 will be obtained. b. Since CCl 4 is the limiting reactant, none of it is left. To calculate the amount of Cr 2 O 3 that is left, we used the limiting reactant to calculate the amount of Cr s O 3 used in the reaction. 2
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This note was uploaded on 02/02/2010 for the course CHM 1220 taught by Professor Barber during the Spring '07 term at Wayne State University.

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Exam I Sample Exam Solutions - Department of Chemistry CHM...

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