# write an equation for the neutralization of h2so4 by koh

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## What Is the Equation for the Neutralization of H2SO4 by KOH?

The formula H2SO4(aq) + 2KOH(aq) –> K2SO4(aq) + 2H2O(l) represents a neutralization reaction of the acidic sulfuric acid and the alkaline potassium hydroxide. When these two chemicals are mixed together, they create a solution of water, or H2O, and potassium sulfate, a salt. Sulfuric acid is a strong acid and potassium hydroxide is a strong base. The reaction that takes place is exothermic; this means that heat is a byproduct of the reaction. write an equation for the neutralization of h2so4 by koh

### Write the balanced molecular equation for the neutralization reaction between H2SO4 and KOH in aqueous solution?

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

#### Explanation:

Alternatively….

Sulfuric acid + potassium hydroxide potassium sulfate + water

In the symbol equation both mass and charge are balanced, as is mandatory.

The reaction is H2SO4 + 2KOH = K2SO4 + 2H2O

Notice how it is balanced so that there are the same number of each type of atom on both sides of the reaction.

The potassium sulfate will ionize into K+ ions and SO4– ions, but the formula is usually written as above.

KOH + H2SO4 –> KHSO4 + H2O
KHSO4 + KOH –> K2SO4 + H2O
As a net reaction,
2KOH + H2SO4 –> 2H2O + K2SO4
As a net ionic equation,
2OH- + 2H+ –> 2H2O

Potassium hydroxide is a strong base; it exists in water as separated potassium ions and hydroxide ions. Sodium sulfate is a salt of a strong base (NaOH) and a strong acid (H2SO4, which in water, essentially loses both hydrogen ion), so that aqueous solutions of sodium sulfate contain sodium ions and sulfate ions. Thus, there is no acid/base reaction between KOH and NaSO4, Furthermore, all potassium and sodium salts are soluble in water so no precipitate forms when solutions of KOH and Na2SO4 are mixed.

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Use the lab equation: V(mL) x Molarity = number of moles

Each mol of H2SO4 reacts with two moles of KOH.

0.050 moles of H2SO4? If so, 0.1 mol of KOH.

With Concentrated sulphuric Acid it is possible As, Copper does not displace hydrogen from non-oxidizing acids like dilute H2SO4. But, concentrated sulfuric acidis an oxidizing agent ;which reacts withcopper to form copper sulfate. copperdoes not react with sulphuric acidbecause it is not reactive enough to do so.

2 KOH + H2SO4 ———→ K2SO4 + 2 H2O

now if we have 10.75g of H2SO4 we can calculate the number of moles = Wt.g/M.wt = 10.75 / 98 = 0.109 mole

now we can calculate the number of moles of K2SO4 as that:

1 mole of H2SO4 react gives 1 mole K2SO4

0.109 mole of H2SO4 gives (x)

x (number of moles of K2SO4 = 1 * 0.109 / 1 = 0.109 mole

now we can calculate the wt.g as wt.g = no. moles * M.wt

wt.g = 0.109 * ((39*2) + 32 +(16*4) = 18.966 g = 19 g write an equation for the neutralization of h2so4 by koh

When determining the limiting reagent, you need to decide which product you will use for comparison. You need to determine which reactant will produce the least number of moles of one of the products. The reactant that produces the least number of moles of the product is the limiting reagent. I’m going to choose H₂ as the product for comparison.

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#### Balanced equation of

2Al(s) + 3H₂SO₄(aq) →Al₂(SO₄)₃(aq) + 3H₂(g)

Calculate moles of Al and H₂SO₄.

moles (n) = mass (m)/molar mass (M).

Moles Al

m(Al) = 10.00 g

M(Al) = 26.982 g/mol

n(Al) = 10.0 g Al/26.982 g Al/mol Al = 0.3706 mol Al

Moles H₂SO₄

m(H₂SO₄) = 30.0 g

M(H₂SO₄) = (2 × 1.008 g/mol H) + 32.06 g/mol S + (4 × 15.999 g/mol O) = 98.07 g/mol

n(H₂SO₄) = 30.0 g H₂SO₄/98.07 g H₂SO₄/mol H₂SO₄ = 0.3059 mol H₂SO₄

Calculate moles of H₂ that can be produced by each reactant, using the mole ratio between each reactant and H₂ in the balanced equation.

Moles Al → Moles H₂

0.3706 mol Al × 3 mol H₂/2 mol Al = 0.5559 mol H₂

Moles H₂SO₄ → Moles H₂

0.3059 mol H₂SO₄ × 3 mol H₂/3 mol H₂SO₄ = 0.3059 mol H₂

Sulfuric acid produces the least number of moles of H₂ , so it is the limiting reagent.