1.

What weight of AgCl will be precipitated whena solution containing 4 . 77 g of NaCl is added to asolution of 5 . 77 g of AgNO_(3) ? (Na = 23, Cl = 35 . 5, Ag = 108 , N = 14 and O = 16)

Answer»

Solution :`NACL + Ag NO_(3) to NA NO_(3) + Ag Cl`
No . Of MOLESOF `NaCl = ( 4 . 77)/( 58 . 5) = 0 . 08154`
Noof moles of `ag NO_(3) = ( 5 . 77) /( 170) = 0 . 0 3394`
Sinceno. of moles of `AgNO_(3)`is lessthenthat of NaClthe whole of `AgNO_(3)` SHALL convertinto AgCl `AgNO_(3)` is a limitingreagent). Applyingthus the POAC forAg atomsas the Agatoms are conserved .
molesofAg in `AgNO_(3) ` = moles of Ag in AgCl
` 1 xx`moles of `AgNO_(3)= 1 xx ` molesof AgCl
`(wt. of AgNO_(3))/( mol. wt. of AgNO_(3)) = (wt.of AgCl)/( mol.wt.of AgCl)`
Wt. of AgCl = ` 0 . 0 3394 xx 143.5 = 4.87` g


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