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 pH = 1. 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

Solution:

CdS burned to CdSO4 and the weight of the same is found to be 0.108 g

CdS + 2O2 = CdSO4

From stoichiometry of the reaction we can write,

When 1 mole CdS bunt then 1 mol CdSO4 forms

Therefore 144 gram CdS burnt then 208.47 gram CdSO4 forms

Since X gram CdS burnt when 0.108 gram CdSO4 found

X = 0.108 * 144/208.47

CdS burnt = 0.07477 gram

secondly when H2S is collected in the sample of CdCl2 some CdS precipitates

CdCl2 + H2S = CdS+ 2HCl

From Stoichiometry when 1 mole H2S is reacted then 1 mole CdS precipitates

34 gram H2S reacts then 144 gram CdS precipitates,

y gram H2S reacts then 0.07477 gram CdS precipates

H2S in the sample = 0.07477*34/144

= 0.01765 gram H2S present in the sample

% H2S in the sample = 0.01765/50 *100

= 0.03530 %

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