Chemistry, 05.02.2020 00:02 mercydiaz84

(a) write the balanced neutralization reaction that occurs between h2so4 and koh in aqueous solution. phases are optional. (b) suppose 0.750 l of 0.480 m h2so4 is mixed with 0.700 l of 0.290 m koh. what concentration of sulfuric acid remains after neutralization?

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Answer from: carlinryan

The answers for the acid-base reaction between sulfuric acid and potassium hydroxide are:

a) The balanced neutralization reaction that occurs between H₂SO₄ and KOH is:

H₂SO₄(aq) + KOH(aq) ⇄ H₂O(l) + K₂SO₄(aq)

b) The concentration of sulfuric acid that remains after the neutralization of 0.750 L of 0.480 M H₂SO₄ with 0.700 L of 0.290 M KOH is 0.108 M.

a) The acid-base reaction between H₂SO₄ and KOH is:

H₂SO₄(aq) + KOH(aq) ⇄ H₂O(l) + K₂SO₄(aq)

To balance the equation, we need to equalize the number of H, S, O, and K atoms in the reactants with the products. On the products side, we have 2 atoms of K, so we need to add a coefficient of 2 before KOH:

H₂SO₄(aq) + 2KOH(aq) ⇄ H₂O(l) + K₂SO₄(aq)

Now, we have 6 atoms of O on the reactants and 5 on the products side, we need to add a coefficient of 2 before H₂O:

H₂SO₄(aq) + 2KOH(aq) ⇄ 2H₂O(l) + K₂SO₄(aq) (1)

Equation (1) is the balanced reaction.

b) The number of moles of acid is equal to the number of moles of the base at neutralization. To find the concentration of the acid, we need to calculate the number of moles of acid and base:

n = C*V

Where:

C: is the molar concentration

V: is the volume

The number of moles of the acid (H₂SO₄):

$n_{a_{i}} = 0.750 L*0.480 mol/L = 0.36 \:moles$

The number of moles of the base (KOH):

$n_{b_{i}} = 0.700 L*0.290 mol/L = 0.203 \:moles$

At the neutralization, we have:

$n_{a} = n_{b}$

$n_{a} = 0.203 \:moles$

Initially, we had 0.36 moles of acid and since 0.203 moles of it reacts with the base, the number of moles of acid that remains after neutralization is:

$n_{f} = 0.36 \: moles - 0.203\: moles = 0.157 \:moles$

Now, the concentration of acid is:

$C_{f} = \frac{n_{f}}{V_{t}}$

Where:

$V_{t}$ : is the total volume = (0.750 + 0.700) L = 1.45 L

$C_{f} = \frac{0.157 \:moles}{1.45 L} = 0.108 M$

Therefore, the concentration of sulfuric acid that remains is 0.108 M.

You can learn more about acid-base reactions here:

linklinklink

I hope it helps you!

Answer from: cocomorillo35181

These are two questions and two answers

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization 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) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

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)

2) Moles of H₂SO₄:

V = 0.750 literM = 0.480 mol/literM = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 literM = 0.290 mol/literM = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

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.

5) Amount of H₂SO₄ that reacts:

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

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 molTotal volume = 0.700 liter + 0.750 liter = 1.450 literConcetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

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