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200cm\(^3\) each of 0.1M solutions of lead (II) trioxonitrate(V) and hydrochloric acid were mixed. Assuming...

Chemistry
JAMB 1999

200cm\(^3\) each of 0.1M solutions of lead (II) trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]

  • A. 2.78g
  • B. 5.56g
  • C. 8.34g
  • D. 11.12g
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Correct Answer: Option A
Explanation

First, we need to write the balanced chemical equation for the reaction;

                           Pb(NO\(_3\))\(_2\)     +       2HCl   →    PbCl\(_2\)   +   2HNO\(_3\)

                            1mole                      2moles

Recall that  n = \(\frac{CV}{1000}\)

                   n =\(\frac{{0.1}\times{200}}{1000}\)

                   n = 0.02 mole of Pb(NO\(_3\))\(_2\)

Similarly, from the question, both HCl and Pb(NO\(_3\))\(_2\) have the same concentration as well as volume. This implies the number of moles of HCl will also be the same. i.e 0.02mole of HCl

However, from the balanced chemical equation, if 2moles of HCl reacted with 1mole of Pb(NO\(_3\))\(_2\)

                               then, from the calculations   0.02 mole HCl will react with 0.01 mole Pb(NO\(_3\))\(_2\) (Based on the stoichiometry above, i.e half the number of moles of HCl)

This implies that HCl will first be used up in the reaction, as the reactant, hence it is called the limiting reagent. Consequently, it will form the basis of our calculation.

The reference will now be between HCl and the insoluble salt, which is PbCl\(_2\).

                                               So, if 2 moles of HCl   →  1 mole of PbCl\(_2\)

                                                     0.02moles HCl     →  0.01mole PbCl\(_2\)

                                                                                       n = \(\frac{mass}{Molar Mass}\)

Molar mass of PbCl\(_2\) = 207 + 2(35.5) = 278g/mol

                                                                                       n = \(\frac{mass}{Molar Mass}\)

                                                                                       m = n x M 

                                                                                       m = 0.01 x 278 = 2.78g


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