Question

How many milliliters of NH3(aq) solution (d = 0.9 g/ml), containing 5% by mass NH3, will be required to precipitate iron as Fe(OH)3 in a 1 g sample that contains 50% by mass Fe20_? Fes-- 3NH, -3H-0-Fe(OH)3 - 3 NH;​

Answer 1

Answer:

The volume of ammonia solution required to precipitate iron as $Fe(OH)_{3}$ is $7.04mL.$

Explanation:

The balanced chemical equation for the precipitation of iron as $Fe(OH)_{3}$ from ammonia solution follows $:$

$Fe^{3+}\ +\ 3NH_{3}\ +\ 3H_{2}O$ →→ $Fe(OH)_{3}\ +\ 3NH_{4}^{+}$

Given that,

The density of ammonia solution $=\ 0.9gml^{-1}$

mass % of the solution $=\ 5$%

  ie, $5\ g$ of $NH_{3}$ in $100g$ of solution

weight of the sample $(Fe_{2}O_{3})$ $=\ 1g$

mass % of $Fe_{2}O_{3}$ in the sample $=\ 50$%

 ie, $50g$ of $Fe_{2}O_{3}$ in $100g$ of sample

To find out the volume of ammonia, first, find out the number of moles of $Fe_{2}O_{3}$ in the sample·

No of moles of $Fe^{3+}$ as  $Fe_{2}O_{3}\ =\ \frac{mass\ of\ Fe_{2}O_{3}}{Molar\ mass\ of\ Fe_{2}O_{3}}$

                                              $=\ \frac{1\ *(\frac{50}{100}) }{160}$

                                              $=\ 3.1\ *\ 10^{-3}$ mol

From the balanced equation, it is clear that in total there are $2$ moles of $Fe^{3+}$ present in the solution·

so,

Total no of moles of  $Fe^{3+} \ =\ 2\ *\ 3.1\ *\ 10^{-3}$

                                               $=\ 6.2\ *\ 10^{-3}$ mol

Next, find out the molarity of ammonia solution·

By using the molarity equation in terms of density,

  M $=\ \frac{density\ *\ weight}{Molar\ mass}$

 $M\ =\ \frac{0.9\ *1000\ *\ (\frac{5}{100} )}{17}$

 $M\ =\ 2.64M$

From the balanced chemical equation,

It is clear that $Fe^{3+}$ reacted with $NH_{3}$ to precipitate $Fe(OH)_{3}$·

So,

⇒ no of equivalent of $Fe^{3+}$  $=$  no of equivalent of $NH_{3}$

⇒ no of equivalent of $Fe^{3+}$ $=$  valency × no of moles

                                        $=\ 3\ *\ 6.2\ *\ 10^{-3}$ eq.

⇒ no of equivalent of $Fe^{3+}$ $=\ 1\ *\ 2.64\ *\ V_{L}$ eq.

On equating these we get,

   ⇒  $3\ *\ 6.2\ *\ 10^{-3}\ =\ 2.64\ *\ V_{L}$

      $V_{L}\ =\ 7.04\ *\ 10^{-3}\\\\ \ \ \ \ \ =\ 7.04mL$

Answer 2

Given:

• Density of $NH_{3}$ (d) = 0.9 g/ml
• Mass percentage of the ammonia solution = 5% ( 5g of $NH_{3}$ in 100g of solution)
• Weight of the sample($Fe_2O_3$)  = 1g
• Mass percentage of $Fe_2O_3$ = 50% (50g  $Fe_2O_3$ in 100g solution)

To Find:

• The amount of ammonia required to precipitate iron into $Fe(OH_{3})$ .

Solution:

• The balanced equation is given as:
• $Fe^{3+}+3NH_3+3H_2O$ →→$Fe(OH_3)+3NH_4^{+}$
• First we should find out no: of moles of $Fe^{3+}$ in $Fe_2O_3$  = $\frac{mass of Fe_2O_3}{Molar mass of Fe_2O_3}$ = $\frac{1*\frac{50}{100} }{160}$ = 0.5/160 = 3.125*$10^{-3}$  mol.
• Total No: of moles of $Fe^{3+}$ = 2*3.125*$10^{-3}$  = 6.25*$10^{-3}$ mol
• Next, we should find the molarity of ammonia solution by which we can find the equivalent moles and volume of ammonia required.
• Molarity equation is given as:
• M = $\frac{density*weight}{molar mass}$ = $\frac{0.9*1000*\frac{5}{100} }{17} = \frac{45}{17} = 2.647 M$
• From the given equation it is clear that iron reacted with ammonia to give us $Fe(OH_{3})$  
• So, No: of equivalent iron = no: of equivalent ammonia = valency*no:of moles = 3*6.25*$10^{-3}$ eq, but it is also = 1*2.647*V
• Equating both the equations we get,
• V = $\frac{3*6.25*10^{-3}}{2.64} = 7.10 ml$

The volume of ammonia required to precipitate iron into $Fe(OH_{3})$ = 7.10 ml.