contestada

Ne double-replacement reaction of iron(ii) nitrate with potassium
ydroxide produces a precipitate of iron(iii) hydroxide.
fe(no3)3(aq) +3koh(aq) +3kno3(aq) + fe(oh)3(s)
that mass of iron(iii) hydroxide is produced when 50.0 g iron(iii)
itrate reacts with excess potassium hydroxide?
a 7.36 g fe(oh)3
w
b 22.1 g fe(oh)3
g
3
c 66.3 g fe(oh)3
d 50.0 g fe(oh)3
check it
skip

Respuesta :

Oseni

The mass of iron (III) oxide that will be produced would be 22.1 grams

Stoichiometric calculation

From the equation of the reaction:

[tex]3KOH + Fe(NO_3)_3 --- > 3KNO_3 + Fe(OH)_3[/tex]

The mole ratio of iron (III) nitrate to iron (III) hydroxide is 1:1.

Mole of 50 g iron (III) nitrate = 50/241.86 = 0.207 moles

Equivalent mole of iron (III) hydroxide = 0.207 moles

Mass of 0.207 moles iron (III) hydroxide = 0.207 x 106.87 = 22.1 grams

More on stoichiometric calculations can be found here: https://brainly.com/question/27287858

#SPJ1

Otras preguntas