Ten dietary iron tablets with a total mass of 11.066 g were ground and mixed thoroughly. Then 2.998 g of the powder were dissolved in HNO3 and heated to convert all iron to Fe3+. Addition of NH3 caused quantitative precipitation of F203.XH2O, which was
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- 5. 0.1500 g sample of chromium ore was dissolved and the chromium oxidized to chromate ion. The solution was treated with 25.00 mL of 0.3500 M AGNO3. The resulting precipitate, from the reaction of the excess and analyte in 1: 1 mole relationship, Ag2CrO4 was removed and discarded. The excess AGNO3 required 30.50 mL of 0.200 M KSCN for titration in a 1:1 mole relationship. Calculate the % Cr2O3 in the ore if the mole relationship between Cr2O3 and CrO42- is 1:2 respectively.2) A 100.0-mL sample of spring water was treated to convert any iron present to Fe2*. Addition of 25.00 mb. of 0.008 M K2Cr207 resulted in the reaction 6F22+ + Cr2O72- + 14H*2 6FE3+ + 2Cr3+ + 14H2O The excess K2C12O7 was back-titrated with 8.53 ml of 0.00949 M Fe2* şolution. Calculate the molar concentration of iron in the sample.A 0.5 g sample of an alloy containing 30% (m/m) iron (MM 55.84 g/mol) is dissolved in acid and diluted to 1 L with deionized water. A 10 mL aliquot was taken for analysis, conveniently diluted and included with an excess of ammonium hydroxide, which led to the formation of a reddish solid. After filtration and washing, the solid was calcined to Fe2O3 (MM 159.69 g/mol). Calculate the mass of this final product.
- A 100.0-mL sample of spring water was treated to convert any iron present to Fe2+.Titration with 35.00-mLof 0.002345 M K2Cr2O7 resulted in the reaction6Fe2+ + Cr2O72- + 14H+ 6Fe3+ 1 2Cr3+ + 7H2OThe excess K2Cr2O7 was back-titrated with 8.32 mL of 0.00897 M Fe2+ solution.Calculate the concentration of iron in the sample in parts per million.3. A 100.0 mL sample of ground water was treated to convert the iron present to Fe2+. Addition of 25.00 mL of 0.002412 M K2Cr2O7 resulted in the reaction below. 6Fe2+ + Cr2O72 - + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O The excess K2Cr2O7 was back-titrated with 8.12 mL of 0.00859 M Fe2+ solution. What is the concentration of Fe in the sample in parts per million?2. 100.0-mL sample of spring water was treated to convert any iron present to Fe2*. Addition of 25.00 mL of 0.002107 M K2Cr207 resulted in the reaction 6FE2+ + Cr20,2- + 14H* ® 6F€³+ + 2Cr3+ + 7H2O The excess K2C1207 was back-titrated with 7.47 mL of a 0.00979 M Fe2+ solution. Calculate iron concentration (molarity) in the sample.
- Chromel is an alloy composed of nickel, iron, and chromium. A 0.6392-g sample was dissolved and diluted to 250.0 mL. When a 50.00-mL aliquot of 0.05173 M EDTA was mixed with an equal volume of the diluted sample, all three ions were chelated, and a 5.15-mL back-titration with 0.06139 M copper(II) was required. The chromium in a second 50.0-mL aliquot was masked through the addition of hexamethylenetetramine; titration of the Fe and Ni required 34.27 mL of 0.05173 M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.0-mL aliquot, and the nickel was titrated with 24.31 mL of the EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy. Percentage of nickel = % Percentage of iron = Percentage of chromium = % 10A sample of iron ore weighing 800 mg was treated with HNO3 , boiled to dryness and redissolved in dilute HCl. After filtration and removal of undissolved silica, the liquid was passed through a Walden reductor. The collected sample was titrated with 0.0210 M KMnO4 , requiring 12.0 mL to reach the end point. Calc %Fe (55.845) in the ore.Determine the percentage of iron in a sample of limonite from the following data:weight of sample = 0.5000g volume of KMNO4 added 50.0mL ; 1.00mL Kmno4 = 0.0005507 g fevolume of FeSO4 used for back titration 8.00mL ; 1.00mL FeSO4= 0.008950 g FeOMW: Fe= 55.85 FeO= 71.84
- A 0.9352g sample of ore containing Fe³+, Al³+ and Sr²+ was dissolved and made up to 500.00 mL. The analysis of metals was performed using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard 0.03145 mol/L EDTA solution, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Given the molar masses: Fe=55.845 g/mol; Al-26.982 g/mol and Sr-87.620 g/mol. a) Determine the percentage of each of the metals in the sample. b) Explain why the change in pH allows the determination of the three ions in this sample.Calcium in a 5.00 mL urine sample was precipitated, redissolved and required 16.15 mL of 0.1157 N KMnO4. Calculate the % (w/v) of calcium in the urine.A student withdraws 3.00Ml of supernate from a saturated solution of KHC8H4O4 at room temp (22C). This sample was titrated to the henolphthalein endpoint and 12.85mL of 0.0997 M NaOH was required A similare tiration was done at 22C for a saturated solution of KHP in 0.500 M KCl. The withgrawn supernant was 5.00mL. The volume of titrant (0.0997M NaOH) used was 11.75. a) calculate the % decrease in solubility by using this equation %decrease = [HC8H4O4]water - [HC8H4O4]sat'd KCl /[HC8H4O4]water