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Estudos Gerais • 03/13/2025

Tratar o carbonato FV=1,06 Na2=2x23/C=12/O=2x16=3x16->106 %F...

Tratar o carbonato FV=1,06 Na2=2x23/C=12/O=2x16=3x16->106 %FAq=1,06*70=74,2 VFAeq=(74,2/100)*1200=890,4 bbl [Na2CO3]=890,4 bbl [NaHCO3]=890,4 bbl Na2CO3 + Ca(OH)2 -> CaCO3 + 2NaOH X=198,6x74/106=138,8 lb:-55,125=2,5 sc/25KG Y=157,6x40/84=75 lb NaOH=75 lb 200 bbl=0,06

Tratar o carbonato FV=1,06 Na2=2x23/C=12/O=2x16=3x16->106 %FAq=1,06*70=74,2 VFAeq=(74,2/100)*1200=890,4 bbl [Na2CO3]=890,4 bbl [NaHCO3]=890,4 bbl Na2CO3 + Ca(OH)2 -> CaCO3 + 2NaOH X=198,6x74/106=138,8 lb:-55,125=2,5 sc/25KG Y=157,6x40/84=75 lb NaOH=75 lb 200 bbl=0,06
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