50 kg of N2 and 10 kg of H2 are mixed to produce Ammonia calculate the moles of ammonia formed identify the limiting reagent in the production of ammonia in the situation
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1N2 +3H2=2NH3
NOW WE WILL SEE WHICH IS L.R
1N2= 50 KA AND I MOLE SO 50/1,50*10^3 GM/1
3H2=10 KG AND 3 MOLE SO 10/3,10*10^3 GM/3,10*10^3/2=5*10^3
SO 2H2 IS LESS SO IT IS L.R
SO WE IGNORE 1N2
AND FOCOUS ONLY 3H2
3 MOLE OF H2................2 MOLE OF NH3
1 MOLE..........................2/3
5*10^3................................5*10^3*2/3=3.30 *10^3 MOLE ABOUT
NOW WE WILL SEE WHICH IS L.R
1N2= 50 KA AND I MOLE SO 50/1,50*10^3 GM/1
3H2=10 KG AND 3 MOLE SO 10/3,10*10^3 GM/3,10*10^3/2=5*10^3
SO 2H2 IS LESS SO IT IS L.R
SO WE IGNORE 1N2
AND FOCOUS ONLY 3H2
3 MOLE OF H2................2 MOLE OF NH3
1 MOLE..........................2/3
5*10^3................................5*10^3*2/3=3.30 *10^3 MOLE ABOUT
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