Question

При реакции нейтрализации было получено 17,4 г сульфата калия. Вычислите количество вещества гидроксида калия взятого для реакции

Answer 1

Если есть какие-то вопросы - пиши в комментарии

Answer 2

Calculating the Amount of Potassium Hydroxide Used in the Neutralization Reaction

To calculate the amount of potassium hydroxide used in the neutralization reaction, we can use the information provided about the amount of potassium sulfate obtained.

Given: - Amount of potassium sulfate obtained = 17.4 g

We need to find the amount of potassium hydroxide used in the reaction.

The balanced chemical equation for the neutralization reaction between potassium hydroxide and sulfuric acid is: 2KOH + H2SO4 → K2SO4 + 2H2O

From the balanced chemical equation, we can see that the molar ratio of potassium hydroxide to potassium sulfate is 2:1. This means that for every 2 moles of potassium hydroxide used, 1 mole of potassium sulfate is produced.

To calculate the amount of potassium hydroxide used, we can follow these steps:

1. Convert the mass of potassium sulfate obtained to moles. 2. Use the molar ratio from the balanced chemical equation to find the moles of potassium hydroxide used. 3. Convert the moles of potassium hydroxide to the amount of substance (in grams).

Let's proceed with the calculations:

1. Convert the mass of potassium sulfate obtained to moles: - The molar mass of potassium sulfate (K2SO4) is 174.259 g/mol. - Using the formula: moles = mass / molar mass - Moles of potassium sulfate = 17.4 g / 174.259 g/mol

2. Use the molar ratio to find the moles of potassium hydroxide used: - From the balanced chemical equation, the molar ratio of potassium hydroxide to potassium sulfate is 2:1.

3. Convert the moles of potassium hydroxide to the amount of substance (in grams): - Using the molar mass of potassium hydroxide (KOH), which is 56.1056 g/mol, we can calculate the amount of potassium hydroxide used.

Let's perform the calculations:

Moles of potassium sulfate = 17.4 g / 174.259 g/mol = 0.1 moles

Moles of potassium hydroxide used = 0.1 moles * 2 = 0.2 moles

Amount of potassium hydroxide used = 0.2 moles * 56.1056 g/mol = 11.22112 g

Therefore, the amount of potassium hydroxide used for the reaction was approximately 11.22 grams.