Question

325 mL of a 0.0700M solution of sodium hydroxide reacts with 75.0 mL of a 0.115 M solution of cobalt (11) sulfate. What mass (ing) of cobalt(II) hydroxide( MM = 109.96) is formed assuming the reaction went to completion? NaOH(aq) + Co2(SO4)3(aq) --> Na2SO4(aq) + CO(OH)3(s) (unbalanced) Write answer to three significant figures. Numeric Response

Answer 1

Number of moles of NaOH = 0.325L * 0.07 mol/L = 0.02275 mol
Number of moles of Co2(SO4)3 = 0.075L*0.115mol/L = 0.008625 mol
Balanced equation:
6NaOH + Co2(SO4)3 ------> 2Co(OH)3 + 3Na2SO4

6 mole of NaOH reacts with 1 mole of Co2(SO4)3.
Number of moles of NaOH required for Co2(SO4)3 = 6*0.008625
                                   = 0.05175 mol
But we only have 0.02275 mol of NaOH. Thus NaOH is the limiting reagent in this reaction.
if 6 mole of NaOH gives 2 moles of Co(OH)3 then 0.02275 mol of NaOH should give = (2*0.02275)/6 = 0.007583 mol of Co(OH)3
Molar mass of Co(OH)3 = 109.96 g/mol
Mass of Co(OH)3 will precipitet    = 0.007583 mol * 109.96 g/mol
                           = 0.833 g
Answer is 0.833 g