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

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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!

User Bie
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