Question #1f782

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Question #1f782

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Answer 1 · 2018-01-29T02:24:48

We will require 23 209 g of $N a B r$ $NaBr$ and 38 321 g of $A g N O_{3}$ $AgNO_3$

Explanation:

93.3 lbs is $93.3 \times 454 = 42358 g$ $93.3 xx 454 = 42 358 g$

which must be converted to moles:

$42358 g \div 187.8 \frac{g}{mol} = 225.55 mol$ $42358 g -: 187.8 g/"mol" = 225.55 "mol"$

(187.8 g is the molar mass of AgBr)

In this quantity, we will have 225.55 mol of $A g^{+}$ $Ag^+$ ions and 225.55 mol of $B r^{-}$ $Br^-$ 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 $A g N O_{3}$ $AgNO_3$ and of $N a B r$ $NaBr$

mass of $A g N O_{3} = 225.55 mol \times 169.9 \frac{g}{mol} = 38321 g$ $AgNO_3 = 225.55 "mol" xx 169.9 g/"mol" = 38321 g$

mass of $N a B r = 225.55 mol \times 102.9 \frac{g}{mol} = 23209 g$ $NaBr = 225.55 "mol" xx 102.9 g/"mol" = 23209 g$

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Question #1f782

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