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A solid mixture 5g consists of lead nitrate and sodium nitrate was heated below $600^{\large\circ}C$ until weight of residue was conotaut.If the loss in weight is 23%,find the amount of lead nitrate and sodium nitrate mixture.

$\begin{array}{1 1}(a)\;3.32g,1.68g&(b)\;2.38g,1.28g\\(c)\;1.14g,1g&(d)\;3.2g,2g\end{array}$

question is wrong it should be 28% as per to the answer
Can you answer this question?
 
 

1 Answer

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Step 1:
$Pb(NO_3)_2\rightarrow PbO+2NO_2+\large\frac{1}{2}$$O_2$
$NaNO_3\rightarrow NaNO_2+\large\frac{1}{2}$$O_2$
$a+b=5$------(1)
The loss in weight for 5g mixture $=5\times \large\frac{28}{100}$$=1.4g$
Residue left =$5-1.4g=3.6g$
Step 2:
The residue contains $PbO+NaNO_2$
331g $pb(NO_3)_2$ gives=223g PbO
$a$g $Pb(NO_3)_2$ gives =$\large\frac{223\times a}{331}$g pbo
Similarly 85g $NaNO_3$ gives =$69g$ $NaNO_2$
$b$g $NaNO_3$ gives =$\large\frac{69\times b}{85}$ g $NaNO_2$
$\large\frac{223\times 9}{331}+\large\frac{69}{85}$$\times b=3.6$-----(2)
Step 3:
From (1) & (2) we get
$a=3.32g$
$b=1.68g$
Hence (a) is the correct answer.
answered Nov 6, 2013 by sreemathi.v
 

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