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Answer 1 · 2018-01-29T02:24:48

We will require 23 209 g of $NaBr$ and 38 321 g of $AgNO_3$

93.3 lbs is $93.3 xx 454 = 42 358 g$

which must be converted to moles:

$42358 g -: 187.8 g/"mol" = 225.55 "mol"$

(187.8 g is the molar mass of AgBr)

In this quantity, we will have 225.55 mol of $Ag^+$ ions and 225.55 mol of $Br^-$ ions.

Since the source of each ion is a salt which also contains one ion per formula unit, we will require 225.55 moles of both $AgNO_3$ and of $NaBr$

mass of $AgNO_3 = 225.55 "mol" xx 169.9 g/"mol" = 38321 g$

mass of $NaBr = 225.55 "mol" xx 102.9 g/"mol" = 23209 g$