Answer:The correct answer is option C.
Explanation:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
At STP, 1 mol of gas occupies 22.4 L of volume
So, 100.0 L volume of [tex]NH_3[/tex]will be occupied by :
[tex]\frac{1}{22.4 L}\times 100.0 L=4.4642 moles[/tex]
According to reaction 2 moles [tex]NH_3[/tex] are obtained from 1 mole of [tex]N_2[/tex].
Then,4.4642 moles of [tex]NH_3[/tex] will be obtained from :
[tex]\frac{1}{2}\times 4.4642[/tex] that is 2.2321 moles of [tex]N_2[/tex]
Mass of [tex]N_2[/tex] gas:
Moles of [tex]N_2[/tex] gas Molar mass of [tex]N_2[/tex] gas:
[tex]2.2321 mol\times 28 g/mol=62.4988\approx 62.50 g[/tex]
Hence, the correct answer is option C.
Answer:
C.62.5 g
Explanation:
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Given, Volume = 100.0 L
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1 atm × 100.0 L = n ×0.0821 L atm/ K mol × 273.15 K
⇒n = 4.4592 moles
From the given reaction,
[tex]N_2_{(g)}+3H_2_{(g)}\rightarrow 2NH_3_{(g)}[/tex]
2 moles of ammonia is formed when 1 mole of nitrogen gas undergoes reaction.
Also, 1 mole of ammonia is formed when 1/2 mole of nitrogen gas undergoes reaction.
4.0446 moles of ammonia is formed when 1/2*4.4592 moles of nitrogen gas undergoes reaction.
Moles of nitrogen = 2.2296 moles
Molar mass of nitrogen, [tex]N_2[/tex] = 28.0134 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]2.2296\ moles= \frac{Mass}{28.0134\ g/mol}[/tex]
Mass = 62.5 g
