A minute sample of AlCl3 is analyzed for chlorine. The analysis reveals that there are 24 chloride ions present in the sample. Find the amount of Aluminium ions and total mass of AlCl3.
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`AlCl_3 rarr Al^(3+)+3Cl^-`
`Al^(3+):Cl^(-) = 1:3`
So we would have 1/3 of `Al^(3+)`than the amount of chloride.
Amount of `Al^(3+) = 1/3*24 = 8`
So we have 8 moles of `Al^(3+)`
`AlCl_3:Cl^(-) = 1:3`
So we would have `1/3` of `AlCl_3`than the amount of chloride.
Amount of `AlCl_3 = 1/3*24 = 8`
So we have 8 moles of` AlCl_3` .
Molar mass of `AlCl_3 = 133.5g`
Mass of `AlCl_3 = 8*133.5 = 1068g`
So total mass of `AlCl_3` in the sample is 1068g.
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