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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

$\begin{array}{1 1}157.8mL\\167.8mL\\147.8mL\\127.8mL\end{array} $

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To calculate the number of moles of the components in the mixture
Suppose $Na_2CO_3$ present in the mixture $x$ g
$NaHCO_3$ present in the mixture $=(1-x)$ g
Molar mass of $Na_2CO_3=23\times 2+12+3\times 16$
$\Rightarrow 106gmol^{-1}$
Molar mass of $NaHCO_3=32+1+12+3\times 16$
$\Rightarrow 84gmol^{-1}$
Moles of $Na_2CO_3$ in $x$ g=$\large\frac{x}{106}$
Moles of $NaHCO_3$ in $(1-x)$ g=$\large\frac{1-x}{84}$
As mixture contains equimolar amounts of the two :
$\large\frac{x}{106}=\frac{1-x}{8.4}$
$84x=106-106x$
$84x+106x=106$
$190x=106$
$x=\large\frac{106}{190}$g=0.558g
Thus,moles of $Na_2CO_3=\large\frac{0.558}{106}$$=0.00526$
Moles of $NaHCO_3=\large\frac{1-0.558}{84}=\frac{0.442}{84}$$=0.00526$
To calculate the moles of HCl required
$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2$
$NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2$
1 mole of $Na_2CO_3$ required HCl =2 moles
0.00526 moles of $Na_2CO_3$ required HCl =$0.00526\times 2=0.01052$ mole
1 mole of $NaHCO_3$ required HCl =1 mole
0.00526 mole of $NaHCO_3$ required HCl =$0.00526$mole
Total HCl required =0.01052+0.00526
$\Rightarrow 0.01578$ moles
To calculate volume of 0.1M HCl
0.1 mole of 0.1M HCl are present in 1000ml of HCl
0.01578 mole of 0.1M HCl will be present in HCl =$\large\frac{1000}{0.1}$$\times 0.01578$
$\Rightarrow 157.8mL$
answered Aug 4, 2014 by sreemathi.v
 

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