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t Vew History Bookmarks Window Help digital wwnorton.com 43% 02/25/18 0 This is a Numeric Entry question / it is worth 1 point / You have unlimited attempts / There is a 1% attempt penalty Question 1 point) a See page 336 1st attempt see Periodic Table 0 See Hint Felll) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(l) to insoluble Fe(Iil): 4Fe(OH)\" (aq) + 4011-(aq) +02(g) + 2H20(,)- 4Fe(OH)3(s) How many grams of O2 are consumed to precipitate all of the iron in 80.0mL of 0.0650 M Fell)? 19,20 > 0

Solution

Moles of Fe(II) = 80 mL x 0.0650 mol/1000 mL => 0.0052 mol

4 moles of Fe(II) reacts with 1 mol of O2

So 0.0052 mol of Fe(II) will require( 0.0052 mol / 4 ) = 0.0013 mol of O2

Molar mass of O2 = 32 g/mol

0.0013 mol x 32 g/mol => 0.0416 g of O2