Asked by Maddy
What mass of precipitate is produced when 40.0 mL 0.100M lead nitrate reacts with 30.0 mL 0.100M aluminum sulfate? Assume the volumes are additive.
I keep getting the answer of 2.7g precipitate formed rather than the answer of 1.21g.
I am using an ice table and find that the moles of Pb(No3)2 is .0004
and moles of Al2(So4)3 is .0003
I then use an ice table and use the limiting reagent as my x value.
I think I am messing up somewhere near the beginning by not remembering what the additive volume does.
Please help!
I keep getting the answer of 2.7g precipitate formed rather than the answer of 1.21g.
I am using an ice table and find that the moles of Pb(No3)2 is .0004
and moles of Al2(So4)3 is .0003
I then use an ice table and use the limiting reagent as my x value.
I think I am messing up somewhere near the beginning by not remembering what the additive volume does.
Please help!
Answers
Answered by
DrBob222
If you show your work, instead of a nebulous instinct of what you may have done wrong, I will find the error for you.
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