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Standard XII

Chemistry

Types of Solution

Calculate the...

Question

Calculate the number of millilitres of $N H_{3} (a q)$ solution (d = $0.986$ g/mL) containing $2.5 %$ by weight $N H_{3}$ , which will be required to precipitate iron as $F e (O H)_{3}$ in a $0.8$ g sample that contains $50 % F e_{2} O_{3}$ (as nearest integer).

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Solution

The precipitation reaction is as follows:
$F e^{3 +} (a q) + 3 N H_{3} (a q) + 3 H_{2} O (l i q) \to F e (O H)_{3} (s) + 3 N H_{4}^{+} (a q)$
Moles of $F e_{2} O_{3}$ in sample $= \frac{0.80 \times 0.5}{160}$ $= 2.5 \times 10^{- 3}$
Moles of $F e^{3 +} (a q) = 2 \times 2.5 \times 10^{- 3} = 5 \times 10^{- 3}$
Molarity of $N H_{3} =$ $M_{N H_{3}} = \frac{0.986 \times 1000 \times 0.025}{17}$ $\Rightarrow M_{N H_{3}} = 1.45$

$3 \times$ moles of $F e^{3 +} =$ moles of $N H_{3}$
Also, moles of $N H_{3}$ $= 1.45 \times V (i n L)$
$V = \frac{3 \times 5 \times 10^{- 3}}{1.45} = 10.34 \times 10^{- 3}$ L or $10.34$ mL

So, the nearest integer is $10$ .

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Calculate the number of millilitres of NH3(aq) solution (d = 0.986 g/mL) containing 2.5% by weight NH3, which will be required to precipitate iron as Fe(OH)3 in a 0.8 g sample that contains 50%Fe2O3 (as nearest integer).

Calculate the number of millilitres of $N H_{3} (a q)$ solution (d = $0.986$ g/mL) containing $2.5 %$ by weight $N H_{3}$ , which will be required to precipitate iron as $F e (O H)_{3}$ in a $0.8$ g sample that contains $50 % F e_{2} O_{3}$ (as nearest integer).