When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:
(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃
Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
Answer:
Explanation:
<u>1) Data:</u>
Base: NaOH
Vb = 15.00 ml = 15.00 / 1,000 liter
Mb = ?
Acid: H₂SO₄
Va = 17.88 ml = 17.88 / 1,000 liter
Ma = 0.1053
<u>2) Chemical reaction:</u>
The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:
- Acid + Base → Salt + Water
- H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)
<u>3) Balanced chemical equation:</u>
- H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)
Placing coefficient 2 in front of NaOH and H₂O balances the equation
<u>4) Stoichiometric mole ratio:</u>
The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:
- 1 mole H₂SO₄ : 2 mole NaOH
<u>5) Calculations:</u>
a) Molarity formula: M = n / V (in liter)
⇒ n = M × V
b) Nunber of moles of acid:
- nₐ = Ma × Va = 0.1053 (17.88 / 1,000)
c) Number of moles of base, nb:
- nb = Mb × Vb = Mb × (15.00 / 1,000)
d) At equivalence point number of moles of acid = number of moles of base
- 0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)
- Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M
I can’t see the question :/
Answer:
Is there any other part to this question? If not I'm pretty sure the answer is 205.5 kJ
Explanation:
