A 1.9300 g impure sample of Fe(NH4)2(SO4)2 (FW = 392.14) was dissolved in 75 mL 0.75 M H2SO4 and was titrated with the standard KMnO4 in problem 1. Titration of the sample required 24.00mL while the blank containing only 75mL 0.75 M H2SO4 required 0.75 mL of the KMnO4 titrant. (Hint: subtract blank). Calculate the concentration of Fe(NH4)2(SO4)2 in %w/w in the impure sample. 5Fe²+ + MnO4 + 8H+ → 5Fe³+ + Mn²+ + 4H₂O
A 1.9300 g impure sample of Fe(NH4)2(SO4)2 (FW = 392.14) was dissolved in 75 mL 0.75 M H2SO4 and was titrated with the standard KMnO4 in problem 1. Titration of the sample required 24.00mL while the blank containing only 75mL 0.75 M H2SO4 required 0.75 mL of the KMnO4 titrant. (Hint: subtract blank). Calculate the concentration of Fe(NH4)2(SO4)2 in %w/w in the impure sample. 5Fe²+ + MnO4 + 8H+ → 5Fe³+ + Mn²+ + 4H₂O
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![A 1.9300 g impure sample of Fe(NH4)2(SO4)2 (FW = 392.14) was dissolved in 75 mL 0.75 M H2SO4 and was
titrated with the standard KMnO4 in problem 1. Titration of the sample required 24.00mL while the blank
containing only 75mL 0.75 M H2SO4 required 0.75 mL of the KMnO4 titrant. (Hint: subtract blank). Calculate the
concentration of Fe(NH4)2(SO4)2 in %w/w in the impure sample.
5Fe²+ + MnO4 + 8H+ → 5Fe³+ + Mn²+ + 4H₂O
8.746% and 8.75% are incorrect](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea1e3b03-1f46-4554-b9d6-7e321b8dca65%2F17adb97a-bc18-4df5-a45a-bb0cc2352ad0%2Fu36ppxg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 1.9300 g impure sample of Fe(NH4)2(SO4)2 (FW = 392.14) was dissolved in 75 mL 0.75 M H2SO4 and was
titrated with the standard KMnO4 in problem 1. Titration of the sample required 24.00mL while the blank
containing only 75mL 0.75 M H2SO4 required 0.75 mL of the KMnO4 titrant. (Hint: subtract blank). Calculate the
concentration of Fe(NH4)2(SO4)2 in %w/w in the impure sample.
5Fe²+ + MnO4 + 8H+ → 5Fe³+ + Mn²+ + 4H₂O
8.746% and 8.75% are incorrect
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