Question

9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of ammonia gas, NH3, measured at room temperature and pressure, is obtained by reacting 0.75 moles of hydrogen with excess nitrogen?

Answer 1

Answer:  3.36 L of ammonia gas

Explanation:

The balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$  

According to stoichiometry :

3 moles of $H_2$ produce = 2 moles of $NH_3$

Thus 0.75 moles of $H_2$ will producee=$\frac{2}{3}\times 0.75=0.50moles$  of $NH_3$

But as percent yield is 30 %, amount of ammonia produced = $\frac{30}{100}\times 0.50moles=0.15moles$

According to ideal gas equation:

$PV=nRT$

P = pressure  = 1 atm

V = Volume  = ?

n = number of moles = 0.15

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$273K$

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

$V=\frac{0.15\times 0.0820 L atm/K mol\times 273K}{1atm}=3.36L$

Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

Answer 2

3.36 L of ammonia gas.

The balanced chemical reaction is:

$N_2(g) + 3H_2(g)\rightarrow 2NH_3(g)$

  According to the above reaction:- :

3 moles of $H_2$ produce = 2 moles of $NH_3$

Thus 0.75 moles of  $H_2$ will produce$\frac{2}{3} \times0.75=0.50\ mol\ NH_3$  

But as percent yield is 30 %, amount of ammonia produced = $\frac{30}{100} \times0.50\ moles\\\\=0.15\ moles$

According to ideal gas equation:

$PV=nRT$

P = pressure  = 1 atm

V = Volume  = ?

n = number of moles = 0.15

R = gas constant =0.0821 Latm\Kmol

T =temperature =273 K

$V=\frac{nRT}{P} \\\\=\frac{0.15\ times0.0820\ Latm\Kmol\times273\ K}{1\ atm} \\\\=3.36\L$

Hence,  3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

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