# In the analysis of 1.5000g sample of feldspar, a mixture of the chlorides of sodium and potassium is obtained which  weighs 0.3450g. ?

Subsequently treatment with AgNO3 furnishes 0.7353 g of AgCI. What g/mol ; NaCI=58.45g/mol ; K2O=94.2g/mol; Na2O=62.00g/mol).

Relevance

Weight of feldspar=1.5 gm

Weight of AgCl= 0.7353 gm

Total weight NaCl+KCl = .3450 gm

First the mass of AgCl will be solved obtained from NaCl and KCl

2NaCl=Na 2 O and 2KCl=K2O

Let the amount of Nacl be x gm

And the amount of KCl is (.3450 - x) gm

NaCl + AgNo3 ---> AgCl + NaNO3

58.5 gm NaCl gives = 143.5 gm AgCl

X gm = 143.5/58.5 into x

Similarly for KCl

KCl + AgNO3 ---> AgCl + KNO3

74.5 gm KCl gives = 143.5gm AgCl

(.3450 - x) gm = 143.5/74.5 x (.3450 - x )

Total weight AgCl (given in the ques) = 0.7353 gm

==>(143.5/58.5) into x + (143.5/74.5) x (.3450 - x)) = 0.7353

When we solve this we get x= 0.1332 gm

Amount of KCl = (0.3450-0.1332) = 0.2118gm

2 NaCl = Na2O

2x58.5 = 62

117gm=62gm

====> 0.1332 = 62/117 x 0.1332

= 0.06262 gm Na2O

%Na2O in 1.5 gm feldspar = 0.06262/1.5 x 100 = 4.17%

Similarly for 2KCl = K2O

2x74.5 = 94

149 gm = 94 gm

149 gm KCl = 94 gm K2O

So ===> 0.2118 = 94/149 x 0.2118 = 0.078 gm K2O

% K2O in 0.5gm feldspar = 0.078/1.5 x 100 = 5.21 % K2O

Na2O= 4.17 %

K2O = 5.21%