Hi!
I think the oxidation state of all the atoms should change. :)
Hope this helps
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
(ANS1)— P4 + 5O2 ---> 2P2O5
(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20
(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4
Answer:
What is the question? I don't understand what you are trying to say.
This element is found in group 3A, period 3
<h3>Further explanation
</h3>
The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
