This is a limiting reagent (LR) problem. You can tell because amounts are give for both reactants.
V is voltage, value, volume. Volume maybe.
........N2 + 3H2 ==> 2NH3
begin volume = 20 + 100 = 120 L.
What volume NH3 will be produced if we use all of the N2 and and excess of H2.
Thats 20L x (2 mols NH3/1 mol N2) = 40 L.
What volume NH3 will be produced if we use all of the H2 and and excess of N2. That's 100 L x (2 mol NH3/3 mol H2) = about 67 L NH3.
You see there are two values for the amount of NH3 which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. That means N2 is the limiting reagent. We won't have any N2 left over unreacted and we will produce 40 L NH3.
How much H2 is left over? How much was used? That's 20 L N2 x (3 mol H2/1 mol N2) = 60 L H2 used. We had 100 initially so we must have 100-60 = 40 L left over. Assuming the pressure doesn't change we will have 40 L H2 which was an excess and we will have 40 L NH3 produced from the N2 so the total is????.
20L of N2 and 100 L of H2 are reacted to produce NH3 what is the decrease in the V of the mixture
N2+3H2--> 2NH3
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