Question

You mix 23.0 mL of 0.225 M FeCl3 with 44.5 mL of 0.461 M NaOH. What mass of Fe(OH)3 (in grams) will precipitate from this reaction mixture? Mass = g One of the reactants (FeCl3 or NaOH) is present in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after Fe(OH)3 has been precipitated? Assume the volumes are additive. Concentration = M

Answer 1

Moles of Fech = Molarity & volime (in litres) = 0.225 X 23x10-3 0 = 5.175 x 10-3 mol Moles of Nash = 0.461 x 44.5x10-3 1 0.021 mol s Fech & 3 NaOH → Feronz + 3 Nach olmole of Fell yields I mole of Felotz 5. 175 x 10-3 mole of felly yields = 5.175x10 3 mol a lo lan ehlou a to of FeCOnda Kown upon the bathrooi = 0.005175 3 mole of Naon yields Imole of Feron I mole of Nadu yields, I mole of Fe(OH)2 0.021 mole of Nash yields, dx0=021 dana 0.007 mole of felony

Page No.: Moles of feloH); produced is less in case of Fe ez Therefore fecl is limiting reagent | Mass of fe coHz produced & Molus x molar Mass = 0·005175 mol X 106.8glne = 0.553 g - 0.553g mole of Fe C needs 3 mole of Noon 0.005175 mole of Fech, needs = 0.00517583 = 0.016 mol of Naoy Moles of Naon left = 0.021-0.016 = 0·005 mol Total volume = 23 ml +44.5 mL 267.5 me Molarity Moles O‘005 67.5X10 -3. Volume (litres) = 0.074 M