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A 1.50 g sample of an ore containing silver was dissolved, and all of the Ag+ was...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 30, 2013 at 1:20 AM via web

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A 1.50 g sample of an ore containing silver was dissolved, and all of the Ag+ was converted to0.124 g of Ag2S. What was the percentage of silver in the ore?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 30, 2013 at 1:35 AM (Answer #1)

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When `Ag^+` was converted to `Ag_2S` we will get the `Ag_2S` precipitate and we can measure the weight of it as it is done here. 

`2Ag^++S^(2-) rarr Ag_2S`

Molar mass of `Ag_2S = 247.8g/(mol)`

Mass of `Ag_2S` obtained `= 0.124g`

Moles of `Ag_2S ` obtained `= 0.124/247.8 = 0.0005`

Mole ratio

`Ag^+:Ag_2S = 2:1`

Amount of Ag in ore `= 0.0005xx2 = 0.001mol`

Molar mass of Ag `= 107.86g/(mol)`

Mass of Ag in ore `= 0.001xx107.86 = 0.1079g`

Percentage of Ag in ore `= 0.1079/1.5xx100% = 7.19%`

So the percentage of Ag in ore is 7.19%

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