Question

What is the volume of NH3 produced in the following reaction when 3.0L of N2 reacts with 4.0L of H2?

Answer 1

Answer:

The volume of NH3 produced is 2.67 L

Explanation:

The chemical equation of the reaction that occurs is:

N₂(g) + 3 H₂(g) --> 2 NH₃(g)

First, the limiting reagent must be determined.  The limiting reagent will be the one that runs out first in the reaction and will limit the amount of product to be obtained.

The reason (ratio) is the comparison of two quantities and two values ​​are measured from the division, then: a / b.

By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), you can see that 1 mole of N₂ will react with 3 moles of H₂.   Then you can say that the ratio between N₂ and H₂ is 1: 3.

Given that the ratio between N₂ and H₂ is 1: 3, to react with 3 L of nitrogen, you would need 9 L of hydrogen (3: 9). But you only have 4 L of hydrogen. Then hydrogen H₂ will be the limiting reagent.

Since the amount of product obtained from the reaction will always depend on the amount of limiting reagent, the relationship between H₂ and NH₃. By stoichiometry, you can see that 3 mole of H₂ will form 2 moles of NH₃.   Then you can say that the ratio between H₂ and NH₃ is 3: 2.

Given that the ratio between H₂ and NH₃ is 3: 2 ( $\frac{volume of H2}{volume of NH3}$ ),  if it reacts 4.0L of H2 then:

$\frac{3}{2} =\frac{4}{volume of NH3}$

$volume of NH3=\frac{4}{\frac{3}{2} }$

$volume of NH3=4*\frac{2}{3}=\frac{8}{3}$

volume of NH₃= 2.67 L

The volume of NH3 produced is 2.67 L

Answer 2

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!