Question

0.9031g of mixture of NaCl and KCl on treatment with conc. $H_{2}SO_{4}$ yields 1.0784g of a mixture of $Na_{2}SO_{4}$ and $K_{2}SO_{4}$. Calculate the % composition of the mixture.

Answer 1

$2NaCl\quad +\quad { H }_{ 2 }{ SO }_{ 4 }\quad \rightarrow \quad { Na }_{ 2 }{ SO }_{ 4 }\quad +\quad 2HCl\quad ..........(i)\\ \quad 117g\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 142g$

$2KCl\quad +\quad { H }_{ 2 }{ SO }_{ 4 }\quad \rightarrow \quad { K }_{ 2 }{ SO }_{ 4 }\quad +\quad 2HCl\quad ..........(ii)\\ 149g\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 174g$

Mass of the mixture = 0.9031 g  

Let the mass of NaCl = x g  

The mass of KCl = (0.9031 - x)) g  

$Mass\quad of\quad { Na }_{ 2 }{ SO }_{ 4 }\quad +\quad { K }_{ 2 }{ SO }_{ 4 }\quad =\quad 1.0784\quad g$

From equation (i)  

$117\quad g\quad NaCl\quad gives\quad =\quad 142\quad g\quad { Na }_{ 2 }{ SO }_{ 4 }\\ \\x\quad g\quad of\quad NaCl\quad will\quad give\quad =\quad \frac { 142\quad x }{ 117\quad g{ \quad of\quad Na }_{ 2 }{ SO }_{ 4 }\quad  }$

From equation (ii)  

149 g of KCl gives 174 g of ${ K }_{ 2 }S{ O }_{ 4 }$

$(0.9031\quad -\quad x)\quad g\quad KCl\quad will\quad give\quad =\quad \frac { 174\quad \times \quad (0.9031\quad -\quad x) }{ 149\quad g\quad of\quad { K }_{ 2 }S{ O }_{ 4 }\quad  }$

$Since,\quad mass\quad of\quad { Na }_{ 2 }S{ O }_{ 4 }\quad and\quad { K }_{ 2 }S{ O }_{ 4 }\quad =\quad 1.0784\quad g$

Therefore,  

$\frac { 142x }{ 117 } \quad +\quad \frac { 174\quad \times \quad (0.9031\quad -\quad x) }{ 149 } \quad =\quad 1.0784\quad g$

On solving, x = 0.518 g  

Mass of NaCl = 0.518 g  

Mass of KCl = 0.9031 - 0.518 = 0.3851 g  

$Percentage\quad of\quad NaCl\quad =\quad \frac { 0.518\quad \times \quad 100 }{ 0.9031 } \quad =\quad 57.36%\\ \\Percentage\quad of\quad KCl\quad =\quad \frac { 0.3851 }{ 0.9031 } \quad \times \quad 100\quad =\quad 42.64%$