Help with Calculating 2 mole N2 + H2 = dcm3NH3

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To calculate the volume of ammonia produced from the reaction of 2 moles of nitrogen (N2) with hydrogen (H2), one must consider the limiting reagent, as not all reactants will convert to product. The synthesis of ammonia follows the equation N2 + 3H2 → 2NH3, indicating that 2 moles of N2 require 6 moles of H2 to produce 4 moles of NH3. The volume of ammonia can be calculated using the ideal gas law (PV=nRT), where the volume is dependent on pressure and temperature. The initial calculation of 44.82 dm3 is incorrect because it does not account for the stoichiometry of the reaction and the limiting reagent. Properly determining the amount of H2 available is crucial for an accurate volume calculation of NH3 produced.
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Homework Statement


How much dcm3NH3would we get if 2 mole of N2 react with the same amount of H2


Homework Equations


I tried this n=V/Vm then get V=n x Vm then get 2 mole x 22.41 dm3 x mole -1then get 44.82 dcm3 but i don't now if it is right so any help would be great


The Attempt at a Solution

 
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If by dcm^3 you mean mass (d=m/v --->dv=m?) then you don't need to do all the stuff you wrote down. You need to write down the chemical reaction because the synthesis of ammonia will have a limiting reagent (I'll let you figure that out for yourself). From there you can figure out how many moles of ammonia are made and just use the molar mass to convert to g/Kg of whatever of ammonia.
 
no i was meaning the volume
 
Well regardless of what you are trying to find you need to start the problem the way I explained above. You will not get 2 mol of ammonia when you mix 2 mol of nitrogen and 2 mole of hydrogen because there will be a limiting reagent.

The volume of the gas will depend on the pressure and temperature of the system and number of mols, the relationship is given by PV=nRT. T is temp in Kelvin, and R is the Rydberg constant (0.08206 L Atm/mol K), P is pressure measured in Atm, V is volume measured in L.
 
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