Question #55368
Calculate the number of milliters of 0.582 M NaOH required to precipitate as Fe(OH)2 all of the Fe2+ ions in 110 mL of 0.451M FeCl2 solution.
Expert's answer
Reaction:
Fe2+ + 2OH- = Fe(OH)2↓
So, n (NaOH) = 2n(Fe2+)
n (Fe2+) = V×c = 0.1 × 0.451 = 0.0451 (mol)
n (NaOH) = n (Fe2+) × 2 = 0.0451 × 2 = 0.0902 (mol)
c = n/V; V = n/c
V(NaOH) = 0.0902 / 0.582 = 0.155 (L)
Answer: V(NaOH) = 155 mL
Fe2+ + 2OH- = Fe(OH)2↓
So, n (NaOH) = 2n(Fe2+)
n (Fe2+) = V×c = 0.1 × 0.451 = 0.0451 (mol)
n (NaOH) = n (Fe2+) × 2 = 0.0451 × 2 = 0.0902 (mol)
c = n/V; V = n/c
V(NaOH) = 0.0902 / 0.582 = 0.155 (L)
Answer: V(NaOH) = 155 mL
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#340153 on Dec 2023
#340153 on Dec 2023