Answer:
The answer to your question is: letter c
Explanation:
Data
V1 = 612 ml n1 = 9.11 mol
V2 = 123 ml n2 = ?
Formula
n2 = 1.83 mol
2NH₂ + O₂ → N₂ + 2H₂O
<u>Explanation:</u>
Balancing the equation means, the number of atoms on both sides of the equation must be the same.
In the case of the given equation, we have to find out whether it is balanced or not.
2NH₂ + O₂ → N₂ + 2H₂O
Atoms Number of atoms before balancing after balancing
LHS RHS LHS RHS
N 1 2 2 2
H 2 2 4 4
O 2 1 2 2
To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.
First the theoretical yield of Nabr
by use of mole ratio between FeBr3 and NaBr which is 2:6 the theoretical yield
=2.36 x6/2= 7.08 moles
the % yield = actual yield/ theoretical yield x 100
that is 6.14/7.08 x100= 86.72%
If you look at AIF3 and AICI3, the F ion is smaller than a CI ion. that’s why AICI3 will make a covalent bond while AIF3 will make a ionic bond. explanation: AICI3 doesn’t have a complete transfer of electrons between the metal and the non-metal.
Answer:
2.61 g of NO will be formed
The limiting reagent is the O₂
Explanation:
The reaction is:
4NH₃ + 5O₂ → 4NO + 6H₂O
We convert the mass of the reactants to moles:
3.25g / 17 g/mol = 0.191 moles of NH₃
3.50g / 32 g/mol =0.109 moles of O₂
Let's determine the limiting reactant by stoichiometry:
4 moles of ammonia react with 5 moles of oxygen
Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent
Ratio with NO is 5:4
5 moles of oxygen produce 4 moles of NO
0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO
We convert the moles to mass, to get the answer
0.0872 mol . 30g / 1 mol = 2.61 g
