The Habor process for making ammonia from nitrogen in the air is given by the eq

Question

The Habor process for making ammonia from nitrogen in the air is given by the equation N2+ 3 H2 2 NH3. Calculate the mass of hydrogen that must be supplied to make 500 kg of ammonia in a system that has an 88.8% yield. Carbon monoxide is the chief waste product of the gasoline combustion process. Auto catalytic converters to convert carbon monoxide into carbon dioxide: 2 CO + O2 2 CO2 When 12.6 grams of carbon monoxide is allowed to react with 5.22 grams of oxygen gas, how many grams of carbon dioxide are formed? What amount of the excess reactant remains after the reaction is complete? manufacturers use 35

Explanation / Answer

actual yield = 500kg = 500000g
Theoretical yield      =
percent yield   = atual yield*100/Theoretical yiled
   88.8         = 500000*100/Theoretical yiled
Theoretical yiled = 500000*100/88.8   = 563063g

N2(g) + 3H2(g) --------->2NH3(g)
2 moles of NH3 decomposes to gives 3 moles of H2
2*17g of NH3 decomposes to gives 3*2g of H2
563063g of NH3 decomposes to gives = 3*2*563063/2*17     = 99364g of H2
mass of H2 = 99364g

2CO + O2------> 2Co2
no of moles of CO = W/G.M.Wt
                   = 12.6/28   = 0.45moles
no of moles of O2 = W/G.M.Wt
                   = 5.22/32   = 0.163moles
2 moles of CO react with 1 moles of O2
0.45 moles of CO react with = 1*0.45/2 = 0.225 moles of O2
O2 is limiting reactant
1 moles of O2 to gives 2 moles of CO2
0.163 moles of O2 to gives = 2*0.163/1   = 0.326 moles of CO2
mass of CO2 = no of moles * gram molar mass
             = 0.326*44   = 14.344g
1 moles of O2 react with 2 moles of CO
0.163 moles of O2 react with = 2*0.163/1 = 0.326 moles of CO
no of moles of excess reactant remains after complete the reaction = 0.326-0.163 = 0.163moles
the amont of excess reactant remains after complete the raction = no of moles* gram molar mass
                                                                 = 0.163*28 = 4.564g

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