- 34- gm of a mixture containing - N - 2- and - H - 2- in - 1-3- by mole is partially converted into - NH - 3- Calculate the vapour density of the mixture -containing remaining - N - 2 - H - 2- -160-and - NH - 3- formed- after reaction if it has been found that the - NH - 3- formed required - 0-5- moles of - H - 3 PO - 4- for complete neutralization-

N_2+3H_2 ⇌ 2NH_3 Let mass of N_2 be x g #160;Then mass of H_2 will be (34 x)g moles of N_2= x28 , moles of H_2= 34 ...

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Byju's Answer

Standard XII

Chemistry

Haber Bosch Process

34 gm of a m...

Question

$34$ gm of a mixture containing $N_{2}$ and $H_{2}$ in $1 : 3$ by mole is partially converted into ${N H}_{3}$ . Calculate the vapour density of the mixture (containing remaining $N_{2}, H_{2}$ and ${N H}_{3}$ formed) after reaction if it has been found that the ${N H}_{3}$ formed required $0.5$ moles of $H_{3} {P O}_{4}$ for complete neutralization.

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Solution

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
Let mass of $N_{2}$ be $x$ $g$
Then mass of $H_{2}$ will be $(34 - x) g$
moles of $N_{2} = \frac{x}{28}$ , moles of $H_{2} = \frac{34 - x}{2}$
as their molar ratio is $1 : 3$
So, $\frac{\frac{x}{28}}{\frac{(34 - x)}{2}} = \frac{1}{3} \Rightarrow \frac{6 x}{28} = 34 - x$
$\Rightarrow 6 x = 34 \times 28 - 28 x$
$\Rightarrow 34 x = 34 \times 28$
$\Rightarrow x = 28 \Rightarrow$ Mass of $N_{2}$
$\Rightarrow 34 - x = 34 - 28 = 6 ⟶$ Mass of $H_{2}$
Number of moles of $N_{2} = \frac{28}{28} = 1$
Number of moles of $H_{2} = \frac{6}{2} = 3$
$N_{2} 1 + 3 H_{2} 3 ⇌ 2 N H_{3} -$
$1 - y$ $3 - 3 y$ $2 y$
As $N H_{3}$ required $0.5$ moles of $H_{3} P O_{4}$
So, $n \times n_{f} (H_{3} P O_{4}) = n \times n_{f} (N H_{3})$
(as $n_{f}$ for $H_{3} P O_{4}$ is $3$ )
$\Rightarrow 0.5 \times 3 = n \times 1$
$n = 1.5$ moles of $N H_{3}$
Thus $2 y = 1.5$
$\Rightarrow y = \frac{1.5}{2} = 0.75$ moles of $N H_{3}$
Number of moles remain in solution Mass remain in solution
$N_{2} ⟶ 1 - 0.75 = 0.25$ $0.25 \times 28 = 7 g$
$H_{2} ⟶ 3 - 3 \times 0.75 = 0.75$ $0.75 \times 2 = 1.5 g$
$N H_{3} ⟶ 2 \times 0.75 = 1.5$ $1.5 \times 31 = 46.5 g$
Vapour density= $\frac{E f f e c t i v e M o l a r m a s s (E M M)}{2}$
$\Rightarrow E M M = \frac{m_{1} + m_{2} + m_{3}}{n_{1} + n_{2} + n_{3}} = \frac{7 + 1.5 + 46.5}{0.25 + 0.75 + 1.5}$
$\Rightarrow E M M = 22$
Vapour density of mixture= $\frac{22}{2} = 11$ .

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34 gm of a mixture containing N2 and H2 in 1:3 by mole is partially converted into NH3. Calculate the vapour density of the mixture (containing remaining N2,H2 and NH3 formed) after reaction if it has been found that the NH3 formed required 0.5 moles of H3PO4 for complete neutralization.