Answer:
Mass = 182.4 g
Explanation:
Given data:
Number of moles of Al₂O₃ = 3.80 mol
Mass of oxygen required = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Now we will compare the moles of aluminum oxide and oxygen.
Al₂O₃ : O₂
2 : 3
3.80 : 3/2×3.80 = 5.7
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 5.7 mol × 32 g/mol
Mass = 182.4 g
Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
- A weak acid and its conjugate base.
- A weak base and its conjugate acid.
Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
A homogeneous mixture, the substances are uniformly distributed throughout the mixture
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
