A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Al(OH)3. The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.107 g. What is the mass percent of Al2(SO4)3 in the sample?

16.2%

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To find the mass percent of Al2(SO4)3 in the sample, we need to know the mass of the compound in the mixture. We can calculate this by subtracting the mass of the Al(OH)3 precipitate from the total mass of the mixture.

Mass of Al2(SO4)3 = Mass of mixture - Mass of Al(OH)3 precipitate
= 1.45 g - 0.107 g
= 1.343 g

Now, we can calculate the mass percent of Al2(SO4)3 in the sample:

Mass percent = (Mass of Al2(SO4)3 / Mass of sample) x 100%
= (1.343 g / 1.45 g) x 100%
โ‰ˆ 92.6%

So, the mass percent of Al2(SO4)3 in the sample is approximately 92.6%.

To find the mass percent of Al2(SO4)3 in the sample, we need to determine how much Al2(SO4)3 was present in the original mixture. We can do this by using stoichiometry and the information provided in the problem.

Let's go through the steps:

1. Calculate the moles of Al(OH)3 precipitate.
We know the mass of the precipitate is 0.107 g. To convert this to moles, we need to divide by the molar mass of Al(OH)3.
The molar mass of Al(OH)3 is:
(2 * molar mass of Al) + (3 * molar mass of O) + (3 * molar mass of H)
= (2 * 26.98 g/mol) + (3 * 16.00 g/mol) + (3 * 1.01 g/mol)
= 101.96 g/mol
Moles of Al(OH)3 = Mass of precipitate / Molar mass
= 0.107 g / 101.96 g/mol

2. Determine the stoichiometric ratio between Al(OH)3 and Al2(SO4)3.
From the balanced chemical equation, we know the stoichiometric ratio between Al2(SO4)3 and Al(OH)3 is 1:2.
This means that for every 1 mole of Al2(SO4)3, 2 moles of Al(OH)3 are produced.

3. Calculate the moles of Al2(SO4)3.
By using the stoichiometric ratio, we can determine the moles of Al2(SO4)3 present in the mixture.
Moles of Al2(SO4)3 = (Moles of Al(OH)3) / 2

4. Finally, calculate the mass percent of Al2(SO4)3.
Mass percent = (Mass of Al2(SO4)3 / Total mass of mixture) * 100
= (Moles of Al2(SO4)3 * Molar mass of Al2(SO4)3 / Mass of mixture) * 100

By plugging in the values we calculated in the previous steps, we can find the mass percent of Al2(SO4)3 in the sample.

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Al2(SO4)3 + 6NaOH ==> 2Al(OH)3 +3Na2SO4

0.107 g Al(OH)3 x [1 mole Al2(SO4)3/2 moles Al(OH)3] = ??g Al2(SO4)3
Convert to percent b7 dividing by 1.45 g and multiplying by 100. .