In the analysis of 0.1 of sample of feldspar, 0.118 g of mixture of NaCl and KCl is obtained which on treatment with `AgNO_3` gives 0.2451 of AgCl. Calculate the precentage of `Na_2O` and `K_2O` in feldspar.

Explanation

Let Feldspar contains x and y g of `Na_2O` and `K_2O` respectively.
`Na_2Oto2NaCloverset(AgNO_3)to2AgCl`
`K_2Oto2KCloverset(AgNO_3)to2AgCl`
`62g Na_2O gives=2xx58.5g of NaCl`
`x g Na_2O gives =(2xx58.5xx x )/(62)=1.8871x g NaCl`
Similarty,
`therefore94gK_(2)Ogives=2xx74.5 g of KCl`
`therefore y of K_(2)O gives=(2xx74.5xxy)/(94)=1.5851 y g KCl`
`therefore 1.887x+1.5851y=0.1180` ....(i)
Also,
`therefore62 g Na_(2)O gives=2xx143.5 g AgCl`
`thereforex g Na_(2) gives=(2xx143.5xx x)/(62)=4.6290x g AgCl`
And,
`therefore94g K_(2)O gives=2xx143.5 g AgCl`
`thereforey g K_(2)O gives=(2xx143.5xxy)/(94)=3.0532 y g AgCl`
`therefore4.6290x+3.0532y=0.2451` ....(ii)
solving equation (i) and (ii), get
`x=0.0179g`
`y=0.0531g`
`therefore% of Na_(2)O=(0.0179xx100)/(0.1)=17.9%`
`% of K_(2)O=(0.0531xx100)/(0.1)=53.1%`