Question

Определить массу воды в реакции нейтрализации между раствором серной кислоты и 250 граммами раствора гидроксида натрия

Answer 1

Calculation of Water Mass in Neutralization Reaction

To determine the mass of water produced in the neutralization reaction between sulfuric acid (H2SO4) and a solution of sodium hydroxide (NaOH), we need to consider the balanced chemical equation for the reaction:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

According to the equation, for every 1 mole of sulfuric acid (H2SO4) and 2 moles of sodium hydroxide (NaOH), 2 moles of water (H2O) are produced.

To calculate the mass of water produced, we need to know the molar mass of water, which is approximately 18.015 g/mol.

Given that we have 250 grams of the sodium hydroxide solution, we need to determine the number of moles of sodium hydroxide present in the solution. To do this, we can use the molar mass of sodium hydroxide, which is approximately 40.00 g/mol.

Using the given information, we can calculate the number of moles of sodium hydroxide:

Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH Number of moles of NaOH = 250 g / 40.00 g/mol

Now that we know the number of moles of sodium hydroxide, we can determine the number of moles of water produced using the stoichiometric ratio from the balanced equation:

Number of moles of H2O = Number of moles of NaOH * (2 moles of H2O / 2 moles of NaOH)

Finally, we can calculate the mass of water produced:

Mass of water = Number of moles of H2O * Molar mass of H2O

Let's calculate the mass of water using the given information.

Calculation:

Given: - Mass of NaOH solution = 250 grams - Molar mass of NaOH = 40.00 g/mol - Molar mass of H2O = 18.015 g/mol

1. Calculate the number of moles of NaOH: - Number of moles of NaOH = 250 g / 40.00 g/mol - Number of moles of NaOH = 6.25 mol 2. Calculate the number of moles of H2O: - Number of moles of H2O = Number of moles of NaOH * (2 moles of H2O / 2 moles of NaOH) - Number of moles of H2O = 6.25 mol * (2 mol / 2 mol) - Number of moles of H2O = 6.25 mol

3. Calculate the mass of water: - Mass of water = Number of moles of H2O * Molar mass of H2O - Mass of water = 6.25 mol * 18.015 g/mol - Mass of water = 112.59 grams

Therefore, the mass of water produced in the neutralization reaction between sulfuric acid and the sodium hydroxide solution is approximately 112.59 grams.