Question

Сколько серной кислоты (в г) потребуется для нейтрализации 24 г гидроксида калия

Answer 1

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Answer 2

Calculation of Sulfuric Acid Required to Neutralize Potassium Hydroxide

To calculate the amount of sulfuric acid (H2SO4) required to neutralize 24g of potassium hydroxide (KOH), we can use the balanced chemical equation for the neutralization reaction between sulfuric acid and potassium hydroxide:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide.

Step 1: Calculate the Moles of Potassium Hydroxide

To find the moles of potassium hydroxide, we can use the formula:

moles = mass / molar mass

The molar mass of potassium hydroxide (KOH) is 56.11 g/mol.

So, moles of KOH = 24g / 56.11 g/mol

Step 2: Calculate the Moles of Sulfuric Acid

Since the molar ratio of sulfuric acid to potassium hydroxide is 1:2, the moles of sulfuric acid required will be half of the moles of potassium hydroxide.

Step 3: Convert Moles to Grams

Finally, we can convert the moles of sulfuric acid to grams using its molar mass, which is 98.08 g/mol.

Let's calculate the moles of potassium hydroxide and then determine the amount of sulfuric acid required.

Calculation:

Moles of KOH = 24g / 56.11 g/mol = 0.428 moles

Moles of H2SO4 required = 0.428 moles / 2 = 0.214 moles

Grams of H2SO4 required = 0.214 moles * 98.08 g/mol = 21.0 grams

Therefore, 21.0 grams of sulfuric acid will be required to neutralize 24 grams of potassium hydroxide.