Answer:
50 because it’s a ratio problem
Explanation:
Na₂CO₃(s) → 2Na⁺(aq) + CO₃²⁻(aq)
The sodium carbonate formed from a strong base and a weak acid. Hydrolysis is subjected to the anion of a weak acid.
CO₃²⁻ + H₂O ⇄ HCO₃⁻ + OH⁻
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻
pH>7 alkaline solution
2Na⁺ + CO₃²⁻ + 2H₂O ⇄ 2Na⁺ + 2OH⁻ + H₂CO₃
There are 3 different elements
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
Manipulated variable: the types of soil
Responding variable: how much t the plants grow
