2o L of N AND 100l OF h REACTED TO PRODUCE NH3 WHAT IS DECREASE IN VOLUME OF MIXTURE

To find the decrease in volume of the mixture, we need to determine the volume of the products formed and subtract it from the total initial volume of the reactants.

From the given information, we have 20 L of nitrogen (N₂) and 100 L of hydrogen (H₂) reacting to produce ammonia (NH₃).

The balanced chemical equation for the reaction is:
N₂ + 3H₂ -> 2NH₃

From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Therefore, the molar ratio between nitrogen and ammonia is 1:2.

First, let's calculate the number of moles of nitrogen and hydrogen present in the reactants:

Moles of N₂ = Volume of N₂ / Molar Volume of N₂ at STP
= 20 L / 22.4 L/mol (molar volume at STP)
= 0.893 moles

Moles of H₂ = Volume of H₂ / Molar Volume of H₂ at STP
= 100 L / 22.4 L/mol (molar volume at STP)
= 4.464 moles

Since the molar ratio of N₂ to NH₃ is 1:2, the moles of NH₃ produced will be twice the moles of N₂ reacted, which is 2 * 0.893 = 1.786 moles.

Now, let's find the volume of NH₃ formed:

Volume of NH₃ = Moles of NH₃ * Molar Volume of NH₃ at STP
= 1.786 moles * 22.4 L/mol (molar volume at STP)
= 40 L

Finally, we can calculate the decrease in volume of the mixture:

Decrease in volume = Total initial volume - Volume of NH₃ formed
= (20 L + 100 L) - 40 L
= 80 L

Therefore, the decrease in volume of the mixture is 80 liters.