please explain steps 11. Nitric acid is manufactured in the Oswald process, whic
ID: 104561 • Letter: P
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please explain steps
11. Nitric acid is manufactured in the Oswald process, which involves the catalytic oxidation of ammonia according to the following equation:s: (i) 4 NH3(g) + 502(g) 4 NO(g) + 6H2O(g) (i) 2 NO(g) + O2(g) 2 NO2(g); (iii) 3 NO2(g) + H2O() 2 HNO3(aq) + NO(g); (NO is fed back into the second reactor) (a) What is the mole fraction of NH, that becomes nitric acid in one reaction cycle? (b) How many kilograms of HNO, will be produced in one reaction cycle from 1.00 metric tonne (1.00 x 103 kg) of ammonia if the reactions have and overall 87.5% yield? (c) If concentrated nitric acid contains 70.0% (by mass) of HNO3 and the solution has density of 1.48 g/mL, how many litters of concentrated nitric acid solution will be produced from this process at the above reaction yield?Explanation / Answer
a) 3*[4NH3 (g)+5O2 (g)--->4NO(g)+6H2O(g)]
6*[ 2NO(g)+O2(g)--->2NO2(g)]
4*[3NO2(g)+H2O(l)--->2HNO3(aq) +NO(g)]
4H2O(g)--->4H2O(l)
--------------------------------------------------------
net rxn: 12NH3 (g)+21O2(g)---->14H2O(g)+8HNO3(aq)+4NO(g)
a)mole fraction of NH3=number of mol of HNO3 formed/ total mol of NH3=8 mol NH3/12mol NH3=2/3
b)mass of NH3=1.0*10^3 kg*(1000g/kg)=10^6 g
mol of NH3=mass/molar mass=10^6g/17.031g/mol=58716.458 mol
mol of HNO3 formed =2/3*58716.458 mol=39144.305 mol
mass of HNO3 formed=39144.305 mol*63.01g/mol=2466482.688 g
But overall yield is 87%
overall mass of HNO3 produced=2466482.688*0.87=2145839.938 g=2145.839 kg=2146 kg=2.15 *10^3 kg
c) Mass of HNO3 produced=2145839.938 g
As the solution to be prepared is 70% by mass ,so the mass going into the solution=2145839.938 g+0.70*2145839.938 g=2145839.938 +1502087.957=3647927.895g
So volume to be prepared=mass/density=3647927.895g/1.48g/ml=2464816.145 ml=2464.816 L=2.465*10^3L
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