If I want to create a solution with #\text{Fe}^{+3}=0.2\ \text{M}# using #5\ g# of #"Fe"_2"O"_3#, how many #"mL"# of water must I add?

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Answer 1 · 2018-04-02T13:53:43

#"156.2 mL"#

Molar mass of #"Fe"_2"O"_3 = "160 g/mol"#

Moles of #"5 g"# of #"Fe"_2"O"_3 = "5 g"/"160 g/mol" = 1/32\ "mol"#

#"Fe"_2"O"_3 -> 2"Fe"^(3+) + 3"O"^(2-)#

#"1 mol"# of #"Fe"_2"O"_3# dissociates to give #"2 mol"# of #"Fe"^(3+)#

#"Molarity" = "Moles of solute"/"Volume of solution (in litres)"#

For #"Fe"^(3+)#

#"0.2 M" = (2 xx 1/32\ "mol")/"V"#

#"V" = 2/32\ "mol" × 1/"0.2 M" = "0.3125 L = 312.5 mL"#