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let 3x mole of Fe react then moles of H2O is 4x m.pdf

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let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g Solution let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g.

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let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g Solution let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g

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let 3x mole of Fe react then moles of H2O is 4x m.pdf

  • 1. let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g Solution let 3x mole of Fe react then moles of H2O is 4x moles Fe3O4 formed=x moles of H2=4x Kc=5.1=x*(4x)^4/((0.1-3x)^3(0.05-4x)^4) solving we get x=0.00572 moles of Fe3O4=0.00572 so mass=moles*mol.wt =0.00572(55.845*3+16*4) =1.32 g