Question

Question #1f782

Answer 1

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

Explanation:

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`