Answer:
[tex]3.8L[/tex] of [tex]NH_{3}[/tex]
Explanation:
First let's write the balanced equation for the reaction :
[tex]N_{2}(g)+3H_{2}(g)[/tex] ⇒ [tex]2NH_{3}(g)[/tex]
By analyzing the coefficients of the reaction we find that the relationship between the compounds is [tex]1:3:2[/tex] (I)
This means that for example with 1 mole of [tex]N_{2}[/tex] and with 3 moles of [tex]H_{2}[/tex] we will obtain 2 moles of [tex]NH_{3}[/tex]
1 mole of gas at STP (Standard Temperature and Pressure) has a volume of 22.4 L
The moles and the volume in liters vary directly therefore we can work with the volume and the relationship (I)
If we divide the volume of [tex]H_{2}[/tex] and [tex]N_{2}[/tex] we obtain :
[tex]\frac{5.7L}{1.9L}=3[/tex] (II)
The number in (II) satisfies the relationship (I) so we can write in terms of liters that :
With [tex]1.9L[/tex] of [tex]N_{2}[/tex] and [tex]5.7L[/tex] of [tex]H_{2}[/tex] (three times more liters that [tex]N_{2}[/tex]) we will produce [tex](1.9L).2=3.8L[/tex] of [tex]NH{3}[/tex] (two times more liters that [tex]N_{2}[/tex]) satisfying the relationship (I)
