One of the steps in the manufacture of nitric acid is the combustion of ammonia with oxygen to produce nitrogen dioxide:
NH3+O2→NO2+H2O
How much nitrogen dioxide (in kg) could be produced from 20,000kg of ammonia if the reaction occured in only 72% yield?
Im stuck on this can anyone help me please :)
Thanks soooooo much!
2007-10-14 09:29:32 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
First you should balance the equation:
4NH3 + 7O2 = 4NO2 + 6H2O
1 mole ammonia produces 1 mole NO2. Weight over MW = Number of moles. Thus:
20,000kg (NH3)/17 = X kg (NO2)/46
1,176.47 moles = X/46
X = 46 x 1,176,47 = 54,117.62 kg's NO2,
This is a perfect yield so, 54,117.62 x 0.72 =
38,964.69 kg actual yield, ANS.
2007-10-14 10:15:50 · answer #1 · answered by Mad Mac 7 · 0⤊ 0⤋
14400kg.
One mole of NH3 makes one mole of NO2 so its a one is to one ration reaction so if it were 100% yield it would be 20000 but its only 72% so 72% of 20000 is 14400.
2007-10-14 16:38:44 · answer #2 · answered by bumblebees6 2 · 0⤊ 0⤋
i got 38,964.71kg
mass = moles x mr
so 20,000/17 = 1176.471
72% = 847.06moles
x (mr of 46) = 38,964.71kg.
I might be wrong but hopefully im not and i helped.
2007-10-14 16:43:26 · answer #3 · answered by Anonymous · 0⤊ 0⤋