What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100 M KOH?
Al(NO3)3 + KOH --------> Al(OH)3 + K(NO3)
So the limiting reagent is the .01 mols Al(NO3)3.
= (.01)(77.98 g/mol Al(OH)3)) = .7798 g
Is this correct?
2007-09-23 13:17:16 · 1 answers · asked by the Jam 2 in Science & Mathematics ➔ Chemistry
No, the answer is not correct.
Reasoning:
Remember that the Law of Conversation of Matter says that matter cannot be created nor destroied. With the equation as written, you are some how magically creating 2 additional hydroxide ions and destroying 2 nitrate ions.
You *must* balance the chemical equation before you do any stoichiometry.
You are correct that you have 0.01 mol Al(NO3)3 to use and you also more likely found that you have 0.02 mol KOH to use.
You must determine which of those two will be used up.
Do this by figuring out how many moles of Al(OH)3 can be produced from both reactant amounts and just pick the lower of the two.
2007-09-23 13:27:01 · answer #1 · answered by lhvinny 7 · 0⤊ 0⤋