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$2.0g$ of a mixture of carbonate ,bicarbonate and chloride of sodium on heating produced 56ml of $CO_2$ at NTP 1.6g of the same mixture required 25ml of 1N HCl solution for neutralization.Calculate percentage of each component present in mixture

$(a)\;20\%,22\%,70\%\qquad(b)\;22\%,23\%,69.4\%\qquad(c)\;21\%,21\%,69.57\%\qquad(d)\;24\%,22\%,68.5\%$

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On heating $NaHCO_3$ decomposes to gives $CO_2$
$2NaHCO_3\rightarrow Na_2CO_3+H_2O+CO_2\uparrow$
Meq of $NaHCO_3$=Meq of $CO_2$
$\large\frac{w_{NaHCO_3}\times 1000}{84}=\frac{56}{22400}$$\times 2\times 1000$
$w_{NaHCO_3}=0.42g$ in 2.0g mixture
$\Rightarrow \large\frac{0.42}{2}$$\times 100$
$\Rightarrow 21\%$
Now suppose in 1.6g mixture,'x'g of Nacl,0.336g(21% of 1.6)$NaHCO_3$ and rest $Na_2CO_3$ then
Wt of $Na_2CO_3=1.6-0.336-x$
$\Rightarrow (1.264-x)g$
Since both $Na_2CO_3% and $NaHCO_3$ react with HCl and therefore
Meq of $Na_2CO_3$+Meq of $NaHCO_3$=Meq of HCl
$\large\frac{1.264-x}{53}$$\times 1000+\large\frac{0.336}{84}$$\times 1000=25\times 1$
$x=0.151$
Weight of NaCl=0.151g and % of NaCl=9.43%
Weight of $NaHCO_3$=0.336g and % of $NaHCO_3$=21.00%
Weight of $Na_2CO_3=1.264-0.151$
$\Rightarrow 1.113$ and % of $Na_2CO_3=69.57\%$
Hence (c) is the correct answer.
answered Jan 7, 2014 by sreemathi.v
 

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