Homework Answers
moles Fe(II) = 0.050 L x 0.0550 M=0.00275
From equation we can see, 4 moles Fe =1 Mole 02, so
moles O2 = 0.0021/4 =0.0006875
now convert moles of O2 to grams:
mass O2 = 0.0006875 mol x 32 g/mol=0.022 g
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Fe(lI) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution,...
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Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III): 4Fe(OH)+ (aq) + 4OH- (aq) + O2 (g) + 2H2O (l) -----> 4Fe(OH)3 (s) How many grams of O2 are consumed to precipitate all of the iron in 60.0 mL of 0.0450 M Fe(II)? ____g
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Feedback See Periodic Table Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(ll) to insoluble Fe(III): 4Fe(OH(aq) +40H (aq) +0,(8) + 2H,00) — 4Fe(OH), ) How many grams of O2 are consumed to precipitate all of the iron in 60.0 mL of 0.0850 MFeClu? 32 1st attempt
See Periodic Table SeeH Fe(ll) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fein to insoluble Fe(lll): 4Fe(OH)*(aq) + 4OH(aq) +0,(8) +2H,0(1) 4Fe(OH),() How many grams of O, are consumed to precipitate all of the iron in 60.0 ml of 0.0850 Mrel)?
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