IN the analysis of `0.5 g` sample of feldspar, a mixture of chlorides of sodium and potassium is obtained which weighs `0.1180 g`. Subsequent treatment of the mixed chlorides with silver nitrate gives `0.2451 g` of silver chloride. What is the percentage of sodium oxide and potassium oxide in the sample ?

Explanation

Suppose the weight of NaCl is x g
`{:(Na_(2)O + K_(2) O overset("Step I") to NaCl + KCI underset(AgNO_(3))overset("Stet II")to AgCl),("(in feldspar) x g (0.1180 - x) 0.2451 g "):}`
Applying POAC for Cl atoms in Step II to calculate x,
moles of Cl in NaCl + moles of Cl in KCI = moles of Cl in AgCl
` 1 xx` moles of NaCl ` + 1 xx ` moles of KCl ` = 1 xx ` moles of AgCl
`(x)/( 58 . 5) + (0.1180)/( 84.5) = (0.2451)/( 143.5) , x = 0.0345 g `
(NaCl = 58.5, KCI = 74. 5 and Ag Cl = 143 . 5 )
`:.` wt. of NaCl = 0.0343 g,
wt. of KCI = 0.0837 g.
Again, applying POAC for Na and K atoms to calculate the weight of `Na_(2)O and K_(2)O` respectively ,
we get,
` 2 xx ` moles of `Na_(2) O` = moles of NaCl
and ` 2 xx` moles of `K_(2)O` = moles of KCI
`:. "wt. of " Na_(2) O = (1)/(2) xx ("wt. of Na Cl")/("mol. wt. of NaCl") xx "mol. wt. of " Na_(2) O `
and wt. of `K_(2) O = (1)/(2) xx ("wt. of KCI")/( "mol . wt. of KCI") xx "mol. wt. of " K_(2)O`
`:. ` wt. of ` Na_(2) O = (1)/(2) xx (0.0343)/( 58.5) xx 62 = 0.018 g `
and wt. of `K_(2) O = (1)/(2) xx (0.0837)/( 74.5) xx 94 = 0.053 g`
`:. % of Na_(2) O = (0.018)/(0.50) xx 100 = 3 . 6 %`
and `% of K_(2) O = (0.053)/( 0.50) xx 100 = 10 . 6% `