A 6.881-g sample containing MgCl_2 and NaCl was dissolved in sufficient water to

Question

A 6.881-g sample containing MgCl_2 and NaCl was dissolved in sufficient water to give a 500-mL volume. Analysis of the chloride content in a 50-0mL aliquot resulted in the formation of 0.5923-g of AgCl. The magnesium in al second 50-0 mL aliquot was precipitated as MgNH_4PO_4; on ignition 0.1796-g of Mg_2P_2O_7 was found. Calculate the percentages of MgCl_2.H_2O and NaCl in the original sample. Mol. Mass: MgP_2O_7 = 222.55 g/mol, NaCl = 58.44 g/mol, MgCl_2.6H_2O = 203.30 g/mol The homogeneity of a standard chloride solution was tested by analyzing portions of the material at the top and bottom with the following found Is homogeneity indicated at the 95% confidence level, use both t-test and f-test. DERIVE and calculate the solution potential for the titration of Sn^2+ with Cr_2O_7^2- at The reaction: 3Sn^2+ + Cr_2O_7^2- + 14H^+ = 2Cr^3+ 3 Sn^4+ + 7 H_2O The hydrogen sulfide, H_2S, in a 50-g sample of crude petroleum was removed by distillation and collected in a solution of CdCl_2. The precipitated CdS was filtered, washed and ignited to CdSO_4. Calculate the percent of H_2S in the sample if 0.108 g of CdSO_4 was recovered ? Mol. Mass: CdSO_4 = 208.47 g/mol and H_2S - 34.08 g/mol

Explanation / Answer

Ans

3Sn2+ + Cr2O72-+ 14H+ = 2Cr3+ + 3Sn4+ + 7H2O

Oxidation half reaction:-

3Sn2+ = 3Sn4+ + 6e-  ; E0 = -0.15 V

Reduction half reaction:-

Cr2O72-+ 6e- ------> 2Cr3+ ; E0 = 1.33 V

Thus, overall reaction:-

3Sn2+ + Cr2O72-+ 14H+ = 2Cr3+ + 3Sn4+ + 7H2O ; E0cell = 1.33 - (-0.15) = 1.48 V

please rate me...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.