Stoichimetry Problem?

Question

Aluminum sulfate is poured into sodium hydroxide solution.
The balanced chemical equation is
Al2(SO4)3+6NaOH → 2Al(OH)3 +3Na2SO4.

If 35.0 mL of 0.400 M (moles) sodium hydroxide is used, how many moles of aluminum hydroxide can be produced? Explain how the aluminum hydroxide could be isolated after the reaction.
If only 0.00442 moles of aluminum hydroxide were isolated, what would the percent yield be?

Answer 1

Atomic weights: Al = 27 S = 32 O = 16 Na = 23 H = 1

Let sodium hydroxide solution be called SH

35.0mLSH x 0.400molNaOH/1000mLSH x 2molAl(OH)3/6molNaOH = (35.0)0.400)(2)/(1000)(6) = 0.00467 mol Al(OH)3

0.00442/0.00467 x 100% = 94.6%

The 35mL is given. The first factor comes from the molarity. The mL SH cancel, leaving the mol NaOH. The second factor comes from the balanced equation. The mol NaOH cancel, leaving mol Al(OH)3.

After precipitating the Al(OH)3, you would filter it away from the supernatant liquid through a fluted circle of filter paper in a funnel.

Answer 2

Here are the concepts, you do the math.

You first need to calculate the grams of NaOH. 0.4000 M solution has 0.4 moles of NaOH in 1 liter. Since the mass of NaoH is 40 grams / mole, multiply this by 0.4 and the solution has 16 grams of NaOH in a liter. But you have only 35 ml, so multiply by .035 and you will have the available grams of NaoH. Now this becomes a standard stoich. problem. Convert the grams of NaOH to moles again. Multiply by the molar ratio of 2 moles of Al(OH)3 for every 6 moles of NaOH (i.e. 1/3). This will give you the answer to part 1. You can campare this to the actual moles isolated and calculate the percent yield.

Al(OH)3 will precipitate, so you can filter it out of the product solution.

Answer 3

6 moles NaOH would yield 2 moles Al2(OH)3

or

1 mole NaOH would yield 1/3 mole Al2(OH)3

35 ml (.4mole/1000ml NaOH ) * 1/3 = 0.00467 moles

To isolate you would dilute with water (bring to Ph 7 with .1M HCl) and filter. Al2(OH)3 is insoluble in H2O. The neutralization would bring the precipitate out of the basic environment created by the NaOH.

%yield = 0.00442/.00467 * 100% = 94.6%