# chemistry

Two solutions were prepared. The first was a 0.672 M solution of aluminum hydroxide and the second was a 0.405 M sulfuric acid solution.
1.) If a total of 500 mL of aluminum hydroxide solution was prepared, how many grams of the solute (AL(OH)3) would have been weighed out in order to make this solution.
2.) how many mL of the aluminum hydroxide solution would be used up if it was reacted with 1182 mL of sulfuric acid solution? How many mL of AL(OH)3 solution remains?
3.) How many grams of each of the products are formed as a result of this chemical reaction?

98.08 g/mol 78.00 g/mol
+
H2SO4 AL(OH)3

-------------------------->

342.15 g/mol 18.02 g/mol
+
AL2(SO4)3 H2O

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1. 2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O

1) How many mols Al(OH)3 do you want? That's M x L = 0.672 x 0.500 = approx 0.3 but you should be more accurate than that.
mols Al(OH)3 = grams/molar mass. You know mols and molar mass; solve for grams.

2)How many mols H2SO4 did you add? That's M x L = 0.405 x 1.182 = approx 0.5
Convert mols H2SO4 to mols Al(OH)3.
Approx 0.5 mol H2SO4 x (2 mols Al(OH)3/3 mol H2SO4) = approx 0.5 x 2/3 = about 0.3 mols Al(OH)3 used. Using M x L = mols, convert 0.26 mols Al(OH)3 to L and convert to mL Al(OH)3 used. Subtract from what you had initially to find mL remaining.

3. This is a limiting reagent problem (LR) and you know that because amounts are given for both reactants. You already know from the problem that the LR is H2SO4. Here is how you do the g H2O produced. I'll leave the Al2(SO4)3 for you.
Convert mols H2SO4 to mols H2O. That's
approx 0.5 mol H2SO4 x (6 mols H2O/3 mols H2SO4) = 0.5 x 6/3 = about 1 mol H2O.
g H2O = mols H2O x molar mass H2O.

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posted by DrBob222

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