B Mass of Cl in 1 L solution 0325 L solution containing 161630g 01L will

# B mass of cl in 1 l solution 0325 l solution

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B) Mass of Cl - in 1 L solution 0.325 L solution containing 16.1630g 0.1L will contain 161630g/.325L*0.1L Mass of Cl - =4.97 of Cl - correct! 13. A student mixes 100.0 mL of 0.500 M AgNO 3 with 100.0 mL of 0.500 M CaCl 2 . (8 points) a. Write the balanced molecular equation for the reaction. b. Write the net ionic equation for the reaction. c.How many grams of precipitate will form? d. What is the concentration of Ag + , NO 3 , Ca 2+ , and Cl in the final solution (assume volumes are additive). 2AgNO3(aq) + CaCl2(aq) ---> 2AgCl(s) + Ca(NO3)2(aq) correct! Ag+ + Cl- --> AgCl correct! moles of AgNO3 = 0.5*0.1 = 0.05 moles moles of CaCl2 = 0.1*0.5 = 0.05 moles 2 moles of AgNO3 reacts with 1 mole CaCl2 so AgCl is limiting reagent moles of CaCl2 used = 0.05/2 = 0.025 moles moles of AgCl formed = moles of AgNO3 consumed = 0.05 moles mass of precipitate = 0.05* 143.32 = 7.166 gm correct! 8 Copyright © 2014 by Thomas Edison State College. All rights reserved. LOVEJOY final total volume = 0.1+0.1 = 0.2 L concentration of: Ag + = 0 M as all AgNO3 is consumed correct! NO3 - = 0.05/(0.1+0.1) = 0.25 M correct! Ca 2+ = 0.05/0.2 = 0.25 M correct! Cl - moles of Cl- = 2*moles of CaCl2 = 2*0.05 = 0.1 moles moles of Cl- consumed as AgCl = 0.05 moles left = 0.1-0.05 = 0.05 Cl - = 0.05/0.2 = 0.25 M correct! 14. Based on oxidation number consideration, explain why carbon monoxide is flammable but carbon dioxide is not. (4 points) (Reference: Chang 4.141) Because flammability is determined by a compounds ability to undergo a combustion reaction; because the C in CO has a lower oxidation state than the C in CO2, CO is more flammable. More specifically, The oxidation number of carbon in CO 2 is 4. This is the maximum oxidation number of carbon. Therefore, carbon in CO 2 cannot be oxidized further, as would happen in a combustion reaction, and hence CO 2 is not flammable. In CO, however, the oxidation number of C is 2. The carbon in CO can be oxidized further and hence CO is flammable. 15. The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.07654 M K 2 Cr 2 O 7 for titration. If the current legal limit of blood alcohol content is 0.08 percent by mass, should the driver be prosecuted for drunken driving? What is the percent alchohol in the driver’s blood? (7 points) 3 C 2 H 5 OH (ethanol) + 2 K 2 Cr 2 O 7 + 8 H 2 SO 4 3 HC 2 H 3 O 2 + 2 Cr 2 (SO 4 ) 3 + 2 K 2 SO 4 + 11 H 2 O 3 C2H5OH (ethanol) + 2 K2Cr2O7 + 8 H2SO4 → 3 HC2H3O2 + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O moles of K2Cr2O7 = 0.07654 x 4.23ml /1000ml= .03237 x 10 -3 moles 2mol K2Cr2O7 required for 3mol C2H5OH 0.3237 x 10 -3 mol k2Cr2O7 =3 mol C2H5OH/2mol K2Cr2O7 x 0.3237x 10 -3 mol K2Cr2O7 9 Copyright © 2014 by Thomas Edison State College. All rights reserved. LOVEJOY = 0.4855 X 10 -3 mol C2H5OH Mass of C2H5OH =moles of C2H5OH x molar mass =0.4855 x10 -3 mol x 46.068 g/mol =0.0233g % by mas of C2H5OH= 0.0233g/10g X 100%= .233% this is over the limit so yes he should be prosecuted. #### You've reached the end of your free preview.

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