(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?

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:

Where:

C: is the molar concentration

V: is the volume

At the neutralization, we have:

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:

Now, the concentration of acid is:

Where:

: is the total volume = (0.750 + 0.700) L = 1.45 L

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

You can learn more about acid-base reactions here:

I hope it helps you!

These are two questions and two answers

    Question 1:

    Question 2:

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:

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

2) Skeleton equation (unbalanced)

#) Balanced chemical equation (including phases)

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

2) Moles of H₂SO₄:

3) Moles of KOH:

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:

         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:

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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

i wish i could but this one is weird give me abc and i can solve it.