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2 years ago

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A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water, and an ex- cess of NaOH is add

ed, 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?

1 answer:

chubhunter [2.5K]2 years ago

5 0

Answer:

The mass percent of aluminum sulfate in the sample is 16.18%.

Explanation:

Mass of the sample = 1.45 g

$Al_2SO_3+6NaOH\rightarrow 2Al(OH)_3+3Na_2SO_4$

Mass of the precipitate = 0.107 g

Moles of aluminum hydroxide = $\frac{0.107 g}{78 g/mol}=0.001372 mol$

According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .

Then 0.001372 moles of aluminum hydroxide will be obtained from:

$\frac{1}{2}\times 0.001372 mol=0.000686 mol$

Mass of 0.000686 moles of aluminum sulfate :

= 0.000686 mol × 342 g/mol = 0.2346 g

The mass percent of aluminum sulfate in the sample:

$=\frac{ 0.2346 g}{1.45g}\times 100=16.18\%$

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Explanation:

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A + B ⇄ C

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= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

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where

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So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

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= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

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if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

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V = 1.64604

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