Question

По уравнению реакции H2SO4 + 2NaOH→Na2SO4 + 2H2O рассчитайте массу серной кислоты, необходимой для полной нейтрализации раствора, содержащего 10г гидроксида натрия.

Answer 1

    хг             10г
H2SO4 + 2NaOH→Na2SO4 + 2H2O
   98             80

n (NaOH) = 10/80=0.125 моль
m (H2SO4) = 0.125*98= 12.25 г

Answer 2

Calculation of the Required Mass of Sulfuric Acid

To calculate the mass of sulfuric acid (H2SO4) required to neutralize a solution containing 10g of sodium hydroxide (NaOH), we need to use the balanced chemical equation for the reaction:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

According to the equation, 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. We can use the molar masses of the compounds involved to convert the given mass of sodium hydroxide to moles, and then determine the corresponding mass of sulfuric acid.

1. Calculate the number of moles of sodium hydroxide (NaOH): - The molar mass of sodium hydroxide (NaOH) is 22.99 g/mol for sodium (Na), 16.00 g/mol for oxygen (O), and 1.01 g/mol for hydrogen (H). Therefore, the molar mass of NaOH is 22.99 + 16.00 + 1.01 = 40.00 g/mol. - To find the number of moles, divide the given mass of sodium hydroxide (10g) by its molar mass: 10g NaOH / 40.00 g/mol NaOH = 0.25 mol NaOH

2. Determine the number of moles of sulfuric acid (H2SO4): - According to the balanced chemical equation, 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Therefore, the number of moles of sulfuric acid is half the number of moles of sodium hydroxide: 0.25 mol NaOH * (1 mol H2SO4 / 2 mol NaOH) = 0.125 mol H2SO4

3. Calculate the mass of sulfuric acid (H2SO4): - The molar mass of sulfuric acid (H2SO4) is 1.01 g/mol for hydrogen (H), 32.07 g/mol for sulfur (S), and 16.00 g/mol for oxygen (O). Therefore, the molar mass of H2SO4 is 1.01 + 32.07 + (4 * 16.00) = 98.09 g/mol. - To find the mass, multiply the number of moles of sulfuric acid by its molar mass: 0.125 mol H2SO4 * 98.09 g/mol H2SO4 = 12.26 g H2SO4

Therefore, the mass of sulfuric acid required to neutralize the solution containing 10g of sodium hydroxide is approximately 12.26 grams