Simple Concentration Stoic Problem?

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SUMMARY

The discussion centers on calculating the mass of aluminum nitrate (Al(NO3)3) required to achieve a nitrate concentration of 0.866 M in 225 ml of water. The correct mass is determined to be 13.8 g, derived from the need for 0.19485 moles of nitrate ions, which corresponds to the stoichiometry of the aluminum nitrate compound. Participants clarify that each mole of aluminum nitrate yields three moles of nitrate ions, leading to the necessity of calculating the molar mass accurately, which is confirmed to be 213 g/mol for Al(NO3)3.

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Lori

Homework Statement



What mass of aluminum nitrate should be dissolved in 225 ml of water to make a solute with a total nitrate concentration of 0.866 M ?

Homework Equations



M = n/L

The Attempt at a Solution


So, i keep getting the wrong answer. THe answer is suppose to be 13.8 g but i get a different number
:

.226 L * (0.866 mols/L) = 0.19485 mols needed

Nitrate molar mass is just 16*3 + 14 + 26.98= 88.98 g/mol

0.19485 * 62 = 17.33 grams of Aluminum nitrate
 
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The chemical formula for aluminum nitrate is ## Al (NO_3)_3 ##. I think that might solve your problem.
 
Charles Link said:
The chemical formula for aluminum nitrate is ## Al (NO_3)_3 ##. I think that might solve your problem.

I get that the molar mass for (NO3)3 is 186 but i still get 40 g of Al(NO3)3
 
Lori said:
I get that the molar mass for (NO3)3 is 186 but i still get 40 g of Al(NO3)3
The molar mass of ## Al (NO_3)_3 ## is 213. Meanwhile, how many moles do you need? With ## (NO3)_3 ## you don't need .195 moles.
 
isnt the nitrate concentration referring to just nitrate?
 
Lori said:
isnt the nitrate concentration referring to just nitrate?
But for every mole of ## Al(NO_3)_3 ## you get 3 moles of ## (NO_3)^- ## in the solution. You do need .195 moles of ## (NO_3)^- ##. That part you got correct.
 
Charles Link said:
The molar mass of ## Al (NO_3)_3 ## is 213. Meanwhile, how many moles do you need? With ## (NO3)_3 ## you don't need .195 moles.
Oh my goodness! I keep forgetting you can just get mole-mole ratio from the molecule itself... thanks!
I spent like so long on this problem ._.
 
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