Given the equation:#N_2+3H_2 rarr 2NH_3#, what volume of #NH_3# at STP is produced if 25 grams of #N_2# is reacted with an excess of #H_2#?

Question

Avery Adams · Accepted Answer

#V = 40 L#

It's interesting to note, in my opinion, that while the reaction cannot occur at standard pressure and temperature, the ammonia that results can be produced in the STP.

Anyhow, #N_2# has molar mass # 14*2=28#, so for 25 g we have

#n_(N_2) = m/(MM) = 25/28 mol#

Two moles of ammonia are produced for every mol of nitrogen, so we have

#n_(NH_3) = 2n_(N_2) = 2*25/28 = 25/14 mol#

Since we know that at STP 1 mol of gas has 22.4 L (this can easily be checked by plugging the appropriate values in #pV = nRT#) We have

#V_(NH_3) = 22.4*25/14 = 40 L#

Luna Chen · Answer

undefined Using the stoichiometry of the balanced chemical equation, we find that 1 mole of N2 produces 2 moles of NH3. From the molar mass of N2, we can determine the number of moles of N2 in 25 grams. Then, using the mole ratio from the balanced equation, we can calculate the number of moles of NH3 produced. Finally, we use the ideal gas law to find the volume of NH3 at STP.

1. Calculate moles of N2: $ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $
2. Use mole ratio to find moles of NH3 produced.
3. Use the ideal gas law: $ \text{PV} = \text{nRT} $ where P = pressure (STP = 1 atm), V = volume, n = moles, R = ideal gas constant, and T = temperature (STP = 273.15 K).

Finally, solve for volume (V).