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A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

Answer» <html><body><p><br/></p>Solution :Suppose, the given mixture <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> x g of lead nitrate.<br/> `therefore` The amount of sodium nitrate in the mixture = `5.<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> - x` g The corresponding equations are: <br/> `2Pb(NO_(3))_(2) to <a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(2 xx 331.22 g)(2PbO) + 4NO_(2) uarr + underset(2 xx 223.2 g)(O_(2)) uarr`<br/> `2NaNO_(3) to underset(2.85 g)(2NaNO_(2)0 + underset(2 xx <a href="https://interviewquestions.tuteehub.com/tag/69-331759" style="font-weight:bold;" target="_blank" title="Click to know more about 69">69</a> g)(O_(2)) uarr` <br/> According to these equations, weight of <a href="https://interviewquestions.tuteehub.com/tag/pbo-597764" style="font-weight:bold;" target="_blank" title="Click to know more about PBO">PBO</a> formed by heating x g of `Pb(NO_(3))_(2)` <br/> `=(2 xx 223.2)/(2 xx 331.22) xx x = 0.6739 g` <br/> and weight of `NaNO_2` obtained by heating `(5.0 -x) g` of `NaNO_(3)` <br/> `=(2 xx 69)/(2 xx 85) xx (5.0-x) = 0.8118(5.0 -x)`g of `NaNO_(3)` <br/> `therefore` Total weight of the residue = = 0.6739 x + 0.8118 (5.0 - x) g <br/> Since, the mixture loses 28% on heating, the residue obtained on heating 5.0 g of mixture <br/> `=(100-28)/100 xx 5.000 = 3.600 g` <br/> The amount of `Pb (NO_3)_2` in the mixture = 3.328 g and the amount of `NaNO_3` in the mixture = 5.0 - 3.328 = 1.672 g</body></html>


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