Question

Срочно!!! При взаимодействии серной кислоты с гидроксидом натрия, масса которого составила 176г, образовался осадок. Найдите массу образовавшегося осадка.

Answer 1

Ответ:

вроде так

H2SO4 + Ba(OH)2 - - - > 2H2O + BaSO4

M(Ba(OH)2) = 34 + 137 = 171

n(Ba(OH)2) = 50 : 171 = 0,3 моль

Соотношение 1 к 1

n(BaSO4) = 0,3 моль

M(BaSO4) = 137 + 64 + 32 = 233

m(ВаSO4) = 233 * 0,3 = 69,9

Округляем до 70

Ответ: 70 грамм

Answer 2

Interaction of Sulfuric Acid with Sodium Hydroxide

When sulfuric acid reacts with sodium hydroxide, a neutralization reaction occurs, resulting in the formation of water and a salt. In this case, the salt formed is sodium sulfate. The balanced chemical equation for this reaction is:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Calculating the Mass of the Precipitate

To calculate the mass of the precipitate formed, we need to consider the stoichiometry of the reaction. From the balanced chemical equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide to produce 1 mole of sodium sulfate.

Given that the mass of sodium hydroxide is 176g, we can calculate the mass of sodium sulfate formed using stoichiometry and molar masses.

1. Calculate the moles of sodium hydroxide: - Moles = Mass / Molar Mass - Moles of NaOH = 176g / (22.99 + 15.999 + 1.008) g/mol - Moles of NaOH = 176g / 39.997 g/mol - Moles of NaOH ≈ 4.401 mol

2. Use the mole ratio from the balanced equation to find the moles of sodium sulfate formed: - Moles of Na2SO4 = Moles of NaOH / 2 - Moles of Na2SO4 ≈ 4.401 mol / 2 - Moles of Na2SO4 ≈ 2.201 mol

3. Calculate the mass of sodium sulfate formed: - Mass = Moles * Molar Mass - Mass of Na2SO4 = 2.201 mol * (22.99 + 32.065 + 4*15.999) g/mol - Mass of Na2SO4 ≈ 2.201 mol * 142.042 g/mol - Mass of Na2SO4 ≈ 312.59g

Answer

The mass of the precipitate formed (sodium sulfate) is approximately 312.59 grams.