The theoretical yield of urea : = 227.4 kg
<h3>Further explanation</h3>
Given
Reaction
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
128.9 kg of ammonia
211.4 kg of carbon dioxide
166.3 kg of urea.
Required
The theoretical yield of urea
Solution
mol Ammonia (MW=17 g/mol)
=128.9 : 17
= 7.58 kmol
mol CO₂(MW=44 g/mol) :
= 211.4 : 44
= 4.805 kmol
Mol : coefficient of reactant , NH₃ : CO₂ :
= 7.58/2 : 4.805/1
=3.79 : 4.805
Ammonia as limiting reactant(smaller ratio)
Mol urea based on mol Ammonia :
=1/2 x 7.58
=3.79 kmol
Mass urea :
=3.79 kmol x 60 g/mol
= 227.4 kg
Answer:
I think it is three times I'm not sure
The answer is decomposition, ocean release and respiration. hope this helps u :D
Data:
n (number of mols) = ?
V (volume) = 2.50 Liters
If:
1 L → 1000 g
2.50 L → y
y = 1000*2.50 = 2500 g
Therefore:
m (mass) = 2500 g
Now:
Molar Mass (MM) of oxygen = 16 g/mol
Formula:

Solving:


Answer:
b Different amounts of food samples were used.
Explanation:
The mass of the two samples needs to be the same in order for the test to be accurate.