A mixture contains only NaCl and Al2(SO4)3. A 1.76-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 0.126 g. What is the mass percent of Al2(SO4)3 in the sample?
%Al2(SO4)3 = (mass Al2(SO4)3/mass sample)*100 = ??
How to find mass Al2(SO4)3.
You have mass Al(OH)3 = 0.126 g. Convert to mass Al2(SO4)3.
How many moles is that?
moles Al(OH)3 = grams/molar mass = ??
There are 2 moles Al in 1 mole Al2(SO4)3; therefore, moles Al2(SO4)3 = moles Al(OH)3 x (1 mole Al2SO4)3/2 moles Al(OH)3) and ?? moles Al(OH)3 x (2/1) = moles Al2(SO4)3.
Convert moles Al2(SO4)3 to grams. g - moles x molar mass and substitute into the top equation.
To find the mass percent of Al2(SO4)3 in the sample, we need to determine the amount of Al2(SO4)3 in the precipitate. We can use the mass of the precipitate and the balanced chemical equation to calculate this.
First, we need to find the moles of Al(OH)3 precipitate. We can use the molar mass of Al(OH)3 to convert the mass of the precipitate to moles.
Molar mass of Al(OH)3:
Al = 26.98 g/mol
O = 16.00 g/mol (x3)
H = 1.01 g/mol (x3)
Molar mass of Al(OH)3 = (26.98) + (3 * 16.00) + (3 * 1.01) = 78.01 g/mol
Moles of Al(OH)3 = mass of precipitate / molar mass of Al(OH)3
= 0.126 g / 78.01 g/mol
= 0.001615 mol
Next, we can determine the moles of Al2(SO4)3 using the balanced chemical equation:
2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O
From the equation, we can see that 2 moles of Al(OH)3 react to form 1 mole of Al2(SO4)3.
Moles of Al2(SO4)3 = 0.001615 mol / 2
= 0.0008075 mol
Finally, we can calculate the mass percent of Al2(SO4)3 in the sample:
Mass percent = (mass of Al2(SO4)3 / mass of sample) * 100
Mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3
= 0.0008075 mol * (2 * (26.98) + 3 * (32.07) + 12 * (16.00))
= 0.42369 g
Mass percent = (0.42369 g / 1.76 g) * 100
= 24.08%
Therefore, the mass percent of Al2(SO4)3 in the sample is 24.08%.
To find the mass percent of Al2(SO4)3 in the sample, we need to determine the moles of Al(OH)3 formed and then use stoichiometry to relate it to the moles of Al2(SO4)3. Here's how you can do it step by step:
1. Start by finding the moles of Al(OH)3 using its molar mass.
- Molar mass of Al(OH)3: (27.0 g/mol x 2) + (16.0 g/mol x 3) + (1.0 g/mol x 3) = 78.0 g/mol
Moles of Al(OH)3 = Mass of precipitate / Molar mass of Al(OH)3
Moles of Al(OH)3 = 0.126 g / 78.0 g/mol
2. Now, use stoichiometry to determine the moles of Al2(SO4)3 that were present in the initial sample.
The balanced equation for the reaction between NaOH and Al2(SO4)3 is:
2 NaOH + Al2(SO4)3 → 2 NaAl(OH)4 + 3 Na2SO4
From the equation, we can see that 2 moles of NaOH react with 1 mole of Al2(SO4)3 to produce 1 mole of Al(OH)3. So,
Moles of Al2(SO4)3 = Moles of Al(OH)3 x (1 mole Al2(SO4)3 / 1 mole Al(OH)3)
3. Calculate the mass percent of Al2(SO4)3 in the sample.
Mass percent of Al2(SO4)3 = (Mass of Al2(SO4)3 / Total mass of the mixture) x 100
The total mass of the mixture given is 1.76 g.
Now, you have all the information needed to solve the problem. Plug in the values and calculate the mass percent of Al2(SO4)3 in the sample.