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
Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169
1 mol = 169 grams.
0.432 mol = x
1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
One would be phosporous whose configuration is 1s2 2s2 2p6 3s2 3p3
answer:
the student <u>who</u> answers the riddle will get the prize
explanation:
- who is the pronoun
- a pronoun is something that substitutes for a noun
Answer:
2B2 + 3O2 → 2B2O3
Explanation:
Balance The Equation: B2 + O2 = B2O3
1. Label Each Compound With a Variable
aB2 + bO2 = cB2O3
2. Create a System of Equations, One Per Element
B: 2a + 0b = 2c
O: 0a + 2b = 3c
3. Solve For All Variables (using substitution, gauss elimination, or a calculator)
a = 2
b = 3
c = 2
4. Substitute Coefficients and Verify Result
2B2 + 3O2 = 2B2O3
L R
B: 4 4 ✔️
O: 6 6 ✔️
hope this helps!
