What is the volume of nh3 produced in the following reaction when 3.0l of n2 reacts with 4.0l of h2

Hey there !

Given the reaction:

N2 + 3 H2 = 2 NH3

At constant pressure and temperature  ,volume is proporcional to moles:

Theoretical moles of  N2 and H2 =>  1:3

Theoretical volume of N2 and H2 => 1:3

Experimental volume of N2 and H2 => 3.0 L  :  4.0 L

0.75 : 1  = 2.25 : 3

Since N2 is in excess reactant   , H2 is the limiting reactant

Therefore:

volume  of NH3 is  2/3  * Volume of H2

= 2/3 * 4.0 = 2.66 L

Hope that helps!

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