These are two questions and two answers
Answer:
Question 1:
- <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>
Question 2:
Explanation:
<u>Question 1:</u>
The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.
The products of an acid-base reaction are salt and water.
This is the sketch of such neutralization reaction:
1) <u>Word equation:</u>
- sulfuric acid + potassium hydroxide → potassium sulfate + water
↑ ↑ ↑ ↑
acid base salt water
<u>2) Skeleton equation (unbalanced)</u>
- H₂SO₄ + KOH → K₂SO₄ + H₂O
<u>#) Balanced chemical equation (including phases)</u>
- H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer
<u>Question 2:</u>
<u>1) Mol ratio:</u>
Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:
- 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)
<u>2) Moles of H₂SO₄:</u>
- V = 0.750 liter
- M = 0.480 mol/liter
- M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol
<u>3) Moles of KOH:</u>
- V = 0.700 liter
- M = 0.290 mol/liter
- M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol
<u>4) Determine the limiting reagent:</u>
a) Stoichiometric ratio:
1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH
b) Actual ratio:
0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH
Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.
<u>5) Amount of H₂SO₄ that reacts:</u>
- Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:
x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒
x = 0.203 / 2 = 0.0677 mol of H₂SO₄
<u>6) Concentration of H₂SO₄ remaining:</u>
- Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol
- Total volume = 0.700 liter + 0.750 liter = 1.450 liter
M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer
at that side where the less number of hydrogen are present.