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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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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`
`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%
Posted by jeew-m on July 30, 2013 at 1:35 AM (Answer #1)
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