1. A 0.897g sample containing chloride ion is treated with excess lead (II) nitr

Question

1.    A 0.897g sample containing chloride ion is treated with excess lead (II) nitrate. The lead (II) chloride precipitated is filtered into a filter paper that weighed 0.923g. After washing and drying the precipitate+ filter paper weighed 2.686g. Calculate the following:

1)    Write the total ionic and net ionic equation for the reaction producing the precipitate:

         total ionic:

            net ionic:

2)    Why was excess lead (II) nitrate used in the above procedure?

3)    moles of lead (II) chloride formed:________

4)    Moles of chloride present in the sample__________ and Mass of chloride present in the sample _________

5)    % of chloride in the sample: ___________

6)    Is the mass of the sample same as the mass of precipitate? _________

Explanation / Answer

1) PbCl2 is the precipitate.

Total ionic is 2Cl- + Pb(NO3)2 ---> PbCl2(ppt.) + 2NO3-

Net ionic is 2Cl- + Pb+ ---> PbCl2(ppt.)

2)excess was used to precipitate each and every chloride ion.

3)total mass of PbCl2=2.686-0.923= 1.763 g

Molecular mass of PbCl2=278 g/mole

so, moles of PbCl2=1.763/278 = 6.342*10^-3.

4)moles of chloride=2*moles of PbCl2=12.684*10^-3

mass of chloride ion=35.5*12.684*10^-3 g

=0.45 g.

5)% chloride =0.45/0.897 *100 = 50.2 %.

6)No.

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