Iron was analyzed by dissolving 1.1324 g sample in concentrated HCL. The solution was diluted in H2O, Fe(III) was precipitated as Fe203.H2O by adding NH3. After filteration, washing the residue ignited to give 0.5394 g of pure Feɔ03 (159.69 g/mol). Calculate %Fe (55.847g/mol) in the sample. O 33.32 % O 45.21 % O 54.21 % O 23.21 % 69.12 %

0.753-g sample of a chlorocarbon compound was analyzed by burning it in oxygen and collecting the evolved gases in a solution of NaOH.  After neutralizing, the sample was treated with 25.74 mL of a 0.264 M AgNO3 solution.  This precipitated the chloride (Cl-) out as AgCl and left an excess of AgNO3.  The excess AgNO3 was titrated with 0.1 M KSCN and required 16.47 mL to reach the endpoint in a Volhard titration. Calculate the % w/w Cl– (35.45 g/mol) in the sample.  Provide your answer to 2 places after the decimal point and without units. Reactions: Cl– + Ag+  →  AgCl(s)                             Reaction 1                   Ag+ + SCN–  →   AgSCN(s)                    Reaction 2 HINT: This is an example of a back-titration.  Steps to success: First, calculate the TOTAL amount (in moles) of Ag+ added from the volume and molarity of the silver nitrate solution. Second, determine the EXCESS amount of Ag+ from the reaction with thiocyanate (Reaction 2 above).  Third, calculate the…

A solution of oxalic acid dihydrate (H₂C₂O₄･2H₂O) with a known concentration of 0.400 M H₂C₂O₄･2H₂O is titrated with a 0.333 M NaO solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H₂C₂O₄･2H₂O, according to the following balanced chemical equation:   H₂C₂O₄･2H₂O + 2NaOH → Na₂C₂O₄ + 4H₂O

Answer the following.