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
The first one is 2
The second is 1
The third is 6
And the fourth is 3
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
You may tell when a solution os formed when the item or particle, such as sugar or salt,
dissolves completely in the solvent, such as water.
Basically, you know when a solution is formed when the material you have placed in the solvent disappears :P
