Question

7 :
0.9031 g of a mixture of NaCl and KCI on treatment with H,SOgave 1.0784 g of a mixtur
Na, so
and K. SO, Calculate percentage composition of the original mixture.​

Answer 1

We need to find the Percentage Composition of the Original Mixture. The Mass % of an element in the Compound is given by ,

$\twoheadrightarrow\boxed{{\sf Mass\% = \dfrac{Mass_{(Element \ in \ 1 \ mole \ of \ Compound)}}{Molar\ Mass_{(Compound)}}\times 100 }}$

• So that , let us take :-

• The amount of NaCl in the mixture be x grams .
• Then , the amount of KCl in the mixture will be ( 0.9031 - x ) gram .

• The equation for 1st reacⁿ is :-

$\sf \to \red{2NaCl }+ H_2SO_4 \longrightarrow\red{ Na_2SO_4} + 2HCl$

• From the above balanced reaction , 2 moles of NaCl equivalent to 117 g reacts to give 142 g of Na₂SO₄ .
• Therefore by Unitary Method , 1 g of NaCl will give 142/117 g of Na₂SO₄ .
• Henceforth , x g of NaCl will give 142x/117g of Na₂SO₄ .

$\boxed{\sf x \ g \ of \ NaCl \ gives \ \purple{\dfrac{142}{117}x \ g}\ of \ Na_2SO_4 .}$

$\rule{200}2$

• The equation for 2nd reacⁿ is :-

$\sf \to \red{ KCl } + H_2SO_4 \longrightarrow\red{ K_2SO_4} + 2HCl$

• From the Reaction , 149 g of KCl gives 174 g of K₂SO₄ .

$\boxed{\sf (0.9031-x)\ g \ of \ NaCl \ gives \ \purple{\dfrac{142}{117}(0.9031-x) \ g}\ of \ K_2SO_4 .}$

$\rule{200}2$

• According to Question :-

$\sf\dashrightarrow \dfrac{142}{117}x +\dfrac{ 174}{149}(0.9031-x) =\dfrac{10784}{10000} \\\\\sf\dashrightarrow 1.21x + (1.67)(0.9031) - 1.67x = 1.0784 \\\\\sf\dashrightarrow 1.21x + 1.50 - 1.67x = 1.07 \\\\\sf\dashrightarrow 0.46x = 0.43 \\\\\sf\dashrightarrow x =\dfrac{43}{46} \\\\\sf\dashrightarrow \boxed{\pink{ \sf x = 0.518 g}}$

$\rule{200}2$

$\boxed{\begin{array}{c} \red{\underline{\sf Percentage \ Composition }} \\\\ \sf \% \ of \ NaCl \ = \ \dfrac{0.518}{0.9031}\times 100 = \pink{ 57.36\%} \\\\ \sf \% \ of \ KCl \ = (100-57.36)\% =\pink{ 42.64\%} \end{array}}$