Answer to Question #310167 in General Chemistry for jay

Question

Considered the reaction:

3 NaOH(aq)+ FeCI3(s) →3 NaCI (aq) + Fe(OH)3 (s)

(molar mass: Fe(OH)3=106.84 g/mole; FeCI3=162.19 g/mol

1) if 300 ml of 0.150 M NaOH were added to a solution containing excess FeCI3, CALCULATE the mass of Fe(OH)3 would be produced

Answer 1

Moles of NaOH used are

(300/1000)×0.15= 0.045moles

Mole fraction= 3:1

Mole of Fe(OH)3 produced are

(1/4)× 0.045= 0.01125moles

$Mass = moles × RMM$

Mass= 0.01125× 106.84

= 1.2g