Question

Определите массу гидроксида натрия, необходимого для нейтрализации 200 г 9,8-ного раствора серной кислоты.

Answer 1

2NaOH + H2SO4 → Na2SO4 + 2H2O
1) m(вещества H2SO4) = 200г * 9,8% : 100% = 19,6г
2) n(вещества H2SO4) = 19,6г : 98г/моль = 0,2моль
3) n(NaOH) = 0,2моль * 2 = 0,4моль
4) m(NaOH) = 0,4моль * 40г/моль = 16г

Answer 2

Calculation of Sodium Hydroxide Mass for Neutralization

To determine the mass of sodium hydroxide needed to neutralize 200 g of a 9.8% sulfuric acid solution, we need to use the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4):

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid. To calculate the mass of sodium hydroxide needed, we can follow these steps:

1. Calculate the number of moles of sulfuric acid: - The concentration of the sulfuric acid solution is given as 9.8%. This means that 100 mL of the solution contains 9.8 g of sulfuric acid. - Since we have 200 g of the solution, the mass of sulfuric acid in the solution is (9.8/100) * 200 = 19.6 g. - The molar mass of sulfuric acid (H2SO4) is 98.09 g/mol. - Therefore, the number of moles of sulfuric acid is 19.6 g / 98.09 g/mol = 0.2 mol.

2. Calculate the number of moles of sodium hydroxide: - From the balanced chemical equation, we know that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid. - Therefore, the number of moles of sodium hydroxide needed is 2 * 0.2 mol = 0.4 mol.

3. Calculate the mass of sodium hydroxide: - The molar mass of sodium hydroxide (NaOH) is 40.00 g/mol. - Therefore, the mass of sodium hydroxide needed is 0.4 mol * 40.00 g/mol = 16 g.

The mass of sodium hydroxide needed to neutralize 200 g of a 9.8% sulfuric acid solution is 16 grams.