How many grams of iron (III) will be produced from the reaction of:?

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How many grams of iron (III) will be produced from the reaction of:?

25.0 mL of 0.10 M iron (III) nitrate with excess sodium hydroxide?

2007-12-09 06:22:46 · 3 answers · asked by Blah 1 in Science & Mathematics ➔ Chemistry

3 answers

no of moles of fe(no3)3=0.025*0.1=2.5*10^-3

therefore no of grams= 56*2.5*10^-3=0.14g

2007-12-09 06:29:33 · answer #1 · answered by lp342 4 · 1⤊ 0⤋

Fe(NO3)3 + 3NaOH ===> Fe(OH)3 + 3NaNO3

Atomic weights: Fe=56 O=16 H=1 Fe(OH)3=107

Let the solution be called S.

25.0mLS x 0.10molFe(NO3)3/1000mLS x 1molFe(OH)3/1molFe(NO3)3 x 107gFe(OH)3/1molFe(OH)3 = 0.27g Fe(OH)3 to two significant figures

2007-12-09 14:31:15 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋

.02782

2007-12-09 14:30:04 · answer #3 · answered by ? 4 · 0⤊ 0⤋