A 1.475-g sample containing NH4Cl(FM 53.492), K2CO3 (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, Na+B(C6H5)4-, to precipitate K+ and NH+4 ions completely:


The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the NH3:
NH+4 + OH- → NH3(g) + H2O
It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of NH4Cl and K2CO3 in the original solid.