(Please explain and show all work as I don't know how to work this one)
A 0.405L solution of Al(OH) 3 is neutralized by 0.200L of a 0.220 M solution of H(NO 3 ). Write
the balanced equation for this reaction and calculate the molarity of the Al(OH) 3 solution.
Thanks!
Can someone please answer as soon as possible, I'm trying to finish but this is my last problem and I cannot figure it out. Thank you.
3HNO3 + Al(OH)3 ==> Al(NO3)3 + 3H2O
mols HNO3 = M x L = 0.220 M x 0.200 L = 0.044 moles HNO3
It will take 1/3 mol Al(OH)3 for every 1 mol HNO3; therefore, 0.044/3 = 0.0147 = moles Al(OH)3 needed.
Then M Al(OH)3 = moles/L = 0.0147 M/0.045 L = ? M
Note: Al(OH)3 is so insoluble I don't know if a solution of that strength can be made but that's the way the problem is solved.