Problems


 * Equation 1) C6H6O + 2 CO2 + 3 NaOH = C8H3O5Na3 + 3 H2O **
 * Equation 2) 2 C8H3O5Na3 + 3 H2SO4 = 2 C8H6O5 + 3 Na2SO4 **

275 g of phenol  2 mole x 40g= 80 g of carbon  C8H3O5NA3 + 3 moles of NaOH + 269.4 x 3 moles= 1098.17 g of sodium hydroxide
 * Site 1- **equation one: 269.4g x 1 + 2 mole x 40g + = C8H3O5NA3 + 3 moles of NaOH + 269.4 x 3 moles= 3.16 g/mole

 phenol: 275 g x $11 = $3025  carbon: 80 g x 47= $560  Sodium hydroxide: 1098.17 x $5=$5490.85  total coast of site one, equation one is $9075.85

 equation two: C8H305Na3 + 1 mole x 269.4 g (sodium hydroxide) + 3 moles x H2SO4 + 300g x 1 mole= 2 C8H605 + 3 x moles Na2S04= 1.05 g/mole

 sodium hydroxide= 269.4 g  sulfuric acid= 3 moles x H2SO4 + 300g x 1 mole= 549.18  carbon= C8H305Na3 x 1 mole= 248.02

 sodium hydroxide = 269.4 x $5 = $1347  sulfuric acid= 549.18 x $3 = $1647.54  carbon: 248.02 x $7 = $1736.14  total cost of site one, equation 2, is $4730.68


 * Site 2- **

** Site 3- ** Equation one: 270g x 1 + 2 mole x 40g + = C8H3O5NA3 + 3 moles of NaOH + 270 x 4 moles= 3.16 g/mole
350 g of phenol 2 mole x 40g= 80 g of carbon C8H3O5NA3 + 4 moles of NaOH + 270 x 3 moles= 822.3 g of sodium hydroxide phenol: 350 g x $11 = $3850 carbon: 80 g x 47= $940 Sodium hydroxide: 822.3 x $5=$4110 total coast of site three, equation one is $8900 Equation two: C8H305Na3 + 1 mole x 269.4 g (sodium hydroxide) + 3 moles x H2SO4 + 379g x 1 mole= 2 C8H605 + 3 x moles Na2S04= 2.01 g/mole sodium hydroxide= 270 g   sulfuric acid= 3 moles x H2SO4 + 379g x 1 mole= 549.18 carbon= C8H305Na3 x 1 mole= 248.02 sodium hydroxide = 270 x $5 = $1350 sulfuric acid= 549.18 x $3 = $1647.54 carbon: 248.02 x $7 = $1736.14 total cost of site three, equation 2 is $4733.71
 * Site 4- **