Why do you think the cold front in figure 1 developed

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

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

3) Moles of KOH:

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

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 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

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

6 0

3 years ago

Which additional product balances the reaction H2SO4 + 2NaOH → Na2SO4 + ? 2H2O 2OH H2O2 H3O

Masteriza [31]

Answer:

a) 2H2O

Explanation:

I took the quiz

4 0

3 years ago

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If 495 milliliters of carbon dioxide at 25°C and 101.3 kilopascals reacts with excess water, what is the theoretical yield of ca

postnew [5]

Answer:- 1.24 g

Solution:- The balanced equation for the formation of carbonic acid by the reaction of carbon dioxide with water is:

$CO_2+H_2O\rightarrow H_2CO_3$

From balanced equation, there is 1:1 mol ratio between carbon dioxide and carbonic acid. So, moles of carbonic acid will be equal to the moles of carbon dioxide used.

Moles of carbon dioxide can be calculated using ideal gas law equation as it's volume, temperature and pressure are given.

we need to convert mL to L, degree C to kelvin and kilopascals to atm.

$495mL(\frac{1L}{1000mL})$ = 0.495 L

25 + 273 = 298 K

$101.3kPa(\frac{1atm}{101.3kPa)})$

= 1 atm

Ideal gas law equation is:

PV = nRT

We want to find out the n, so let's rearrange this:

$n=\frac{PV}{RT}$

R is the universal gas constant and it's value is $\frac{0.0821atm.L}{mol.K}$ .

Let's pug in the values in the equation and solve it for n.

$n=\frac{1*0.495}{0.0821*298}$

n = 0.02 mol

To convert the moles to grams we multiply the moles by the molar mass of carbonic acid.

Molar mass of carbonic acid = 2(1.008)+12.01+3(16.00)

= 2.016+12.01+48.00

= 62.03 gram per mol (rounded to two decimal places)

Let's multiply the moles by molar mass:

$0.02mol(\frac{62.03g}{mol})$

= 1.24 g

So, the theoretical yield of carbonic acid is 1.24 g.

6 0

3 years ago

Read 2 more answers

Which of the following salts are soluble?

Alex787 [66]

Answer to the Question

C. Barium phosphate

8 0

3 years ago

The process of cell division that creates sex cells in sexually reproducing organisms is called

Mashcka [7]

The answer is meiosis. Not Mitosis

7 0

3 years ago

Read 2 more answers