Asked by Anonymous
What mass of solid aluminum hydroxide can be produced when 50.0mL of .200 M Al(NO3)3 is added to 200.0 mL of .100 M KOH
Answers
Answered by
DrBob222
Write the equation and balance it.
Al(NO3)3 + 3KOH ==> 3KNO3 + Al(OH)3
Calculate mols Al(NO3)3 by M x L = mols.
Calculate mols KOH by M x L = mols.
Using the equation, calculate how much Al(OH)3 can be produced by EACH reagent if you had all of the OTHER reagent you needed. The smaller of the two amounts (in mols) will be the amount produced.
Then mols x molar mass = g Al(OH)3.
Post your work if you get stuck.
Check my work.
Al(NO3)3 + 3KOH ==> 3KNO3 + Al(OH)3
Calculate mols Al(NO3)3 by M x L = mols.
Calculate mols KOH by M x L = mols.
Using the equation, calculate how much Al(OH)3 can be produced by EACH reagent if you had all of the OTHER reagent you needed. The smaller of the two amounts (in mols) will be the amount produced.
Then mols x molar mass = g Al(OH)3.
Post your work if you get stuck.
Check my work.
Answered by
J
0.520 g
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