Answer: D. 1600 L
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}[/tex]
[tex]\text{Number of moles of nitrogen}==\frac{1000L}{22.4L}=44.6moles[/tex]
[tex]\text{Number of moles of hydrogen}==\frac{2400L}{22.4L}=107.1moles[/tex]
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
According to stoichiometry :
3 moles of [tex]H_2[/tex] require 1 mole of [tex]N_2[/tex]
Thus 107.1 moles of [tex]H_2[/tex] will require=[tex]\frac{1}{3}\times 107.1=35.7moles[/tex] of [tex]N_2[/tex]
Thus [tex]H_2[/tex] is the limiting reagent as it limits the formation of product and [tex]N_2[/tex] is the excess reagent.
As 3 moles of [tex]H_2[/tex] give = 2 moles of [tex]NH_3[/tex]
Thus 107.1 moles of [tex]H_2[/tex] give =[tex]\frac{2}{3}\times 107.1=71.4moles[/tex] of [tex]NH_3[/tex]
Volume of [tex]NH_3=moles\times {\text {Molar volume}}=71.4moles\times 22.4L/mol=1600L[/tex]
Thus volume of ammonia produced is 1600 L
