Answer:
2 NaI (aq) + Br2 (aq) → 2 NaBr (aq) + I2 (s)
This is an oxidation-reduction (redox) reaction:
2 Br0 + 2 e- → 2 Br-I
(reduction)
2 I-I - 2 e- → 2 I0
(oxidation)
Explanation:
Carbon dioxide is a polar molecule whose positive center is on the carbon atom: This positive center is able to attract (and accept) the lone electron pairs present on the oxide ion (O2-). carbon dioxide is acts as a Lewis acid
A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid. A Lewis acid is defined as an electron-pair acceptor.
In CO molecule, there is a lone pair on both carbon and oxygen. The substance which can donate an electron pair are called Lewis base. It is clear that CO molecule can donate an electron pair and hence, it is a Lewis base. Also, CO can be BOTH a Lewis acid and base.
Oxygen is a Lewis base (that too a weak one), not a Lewis acid. REASON: It has lone pair of electrons, which can be donated to electron-deficient species (Lewis acids).
Methane is Neither a Lewis Acid or Lewis Base.
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer:
They would equally balance because of the chemicles.
Explanation:
