Asked by omar
What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M aluminum nitrate is added to 200.0 mL of 0.100M potassium hydroxide?
Answers
Answered by
DrBob222
Step 1. Write the equation and balance it.
Al(NO3)3 + 3KOH --> Al(OH)3 + 3KNO3
Step 2. Convert what you have to mols.
mols = M x L
mols Al(NO3)3 = 0.2 x 0.05 = ??
mols KOH = 0.1 x 0.200 = ??
Step 3. Determine the limiting reagent (that looks like it is KOH but check me out on that).
Step 4. Determine mols Al(OH)3.
Step 5. Convert mols to grams.
grams = mols x molar mass.
Post your work if you get stuck AND explain in detail what you don't understand.
Al(NO3)3 + 3KOH --> Al(OH)3 + 3KNO3
Step 2. Convert what you have to mols.
mols = M x L
mols Al(NO3)3 = 0.2 x 0.05 = ??
mols KOH = 0.1 x 0.200 = ??
Step 3. Determine the limiting reagent (that looks like it is KOH but check me out on that).
Step 4. Determine mols Al(OH)3.
Step 5. Convert mols to grams.
grams = mols x molar mass.
Post your work if you get stuck AND explain in detail what you don't understand.
Answered by
ngan
.520g of Al(OH)3
Answered by
popbob
git griefed
Answered by
bobpop
Sorry, I think it is 5.20g.
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