HW4 Solutions

HW4 Solutions - 9.7 Basis: Data given in the problem Massof...

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9.7 Basis: Data given in the problem Massof hydrate 10.407g Massof dry sample .9.520g Massofwater 0.887g 9.520 g Bal,ll g mol Bal, =0.0243gmolBal, 391 g Bal, 0.887 g H,O \ 1 g mol H,O = 0.0493 g mol H,O 18.02 g H,O 0.0243 = 0.49. or 2 H,O for 1 Bal,. Thus the hydrate is Bal, .2H,0. 0.0493 . - I Yes, the equation is balanced. b) Basis: 1.5kgZnO 1.5 kg ZnOllOOOg ZnO I gmol Z . no l l gmOlDEZI1214 gDEZ 1.0 kg ZnO 81.40 g ZnO I gmol ZnO I gmol DEZ 1.0 kg DEZ = 12.27kg DEZI x 1000 g DEZ c) 20 em) H20~ 1 1.0 gmol H2° 1 1 gmol DEZ I 123.4 g DEZ ~ 18.0 g H2O 5 gmol HP 1 gmol DEZ =127.4 g DEZI - 9.13 The reactionis C4HlO+ 61°2 -+ 4C02 + 5H20 Basis: I molCJlIO Minimum02 =(LFL) ( mol 0, for complete combustion ) mol C.HIO :;; (1.9%) (6.5/1) = 112.4%1 9.11 a) LHS RHS C 4 4 H 10 10 Zn I I ° 14 14 .
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9.27 Reaction on which to base excess is Fez03 + 3C ~ 2Fe + 3CO Basis: I ton Fez03 a. Theoretical C = I ton Fep12000 IbFeZ03 1 1IbmolFep3 1ton Fep3 159.71bFeZ03 600-451 x 100-133% excess CI 451 159.71bFep3 lib molFeZ03 = 1720 Ib F ez03 reacts .l12!l x 100 = 186.0% FeZ031 2000 Total(35.62IbCO+ 902.4lb CO)= 1938Ib col
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HW4 Solutions - 9.7 Basis: Data given in the problem Massof...

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