D---Ammonia (NH3) is produced from the reaction of molecular nitrogen (N2) and m
ID: 924029 • Letter: D
Question
D---Ammonia (NH3) is produced from the reaction of molecular nitrogen (N2) and molecular hydrogen (H2). What is the maximum volume of ammonia that can be produced from 1.00 kg of N2 and 0.500 kg H2at 1.00 atm and 25.00oC?
E--A 5.00 g sample of aluminum metal and a 10.0 g sample of iron metal are heated to 100oC in a boiling water bath. The metals are then quickly transferred to a beaker containing 97.3 grams of water at 22oC. What is the final temperature of the water-metal mixture, assuming no heat loss to the surroundings?Useful information: specific heat capacity of Al = 0.89 J/g•oC; specific heat capacity of Fe = 0.45 J/g•oC.
Explanation / Answer
D. N2+3H2 ---> 2NH3
nO of mol of N2 = 1*10^3 / 28 = 35.71 mol
No of mol of H2 = 0.5*10^3/2 = 250 mol.
From equation,
1 mol N2 = 3 mol H2 = 2 mol NH3.
Limiting reagent is N2
nO of mol of NH3 PRODUCED = 35.71*2 = 71.42 mol
mass of NH3 produced = 71.42*17 = 1214.14 grams
Volume of NH3 produced = nRT/P = 71.42*0.0821*298/1
= 1747.35 L
E. q lost by metals = q gained by water
mAl*s*DT + mFe*s*DT = mwater*s*DT
5*0.89*(100-T)+10*0.45*(100-T) = 97.3*4.18*(T-22)
T = final temperature of mixture = 23.7 C
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