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 reaction is properly written as
Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s)
Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol
Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429
Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,
Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g
Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
The focal length, like you said it's the distance between the FOCAL point and the mirror.
Hope this helps.... :)
Answer:
see the pic for the answer
