Question

Серная кислота массой 17 грамм реагирует с гидроксидом бария. Какова масса осадка?

Answer 1

Ba(OH)2 + H2SO4 = BaSO4(осадок) + 2HOH
M(BaSO4) = 137+32+16*4=233 г/м
M(H2SO4) = 2+32+16*4=98 г/м
m(BaSO4) = 17*233/98 = 40.4 г

Answer 2

Calculation of the Mass of Precipitate

To determine the mass of the precipitate formed when sulfuric acid (H2SO4) with a mass of 17 grams reacts with barium hydroxide (Ba(OH)2), we need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between sulfuric acid and barium hydroxide is:

H2SO4 + Ba(OH)2 → BaSO4 + 2H2O

From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 1 mole of barium hydroxide to produce 1 mole of barium sulfate and 2 moles of water.

To calculate the mass of the precipitate (barium sulfate), we need to determine the number of moles of sulfuric acid and barium hydroxide involved in the reaction.

1. Calculate the number of moles of sulfuric acid: - The molar mass of sulfuric acid (H2SO4) is 98 g/mol. - The mass of sulfuric acid given is 17 grams. - Therefore, the number of moles of sulfuric acid is calculated as: moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4 moles of H2SO4 = 17 g / 98 g/mol

2. Calculate the number of moles of barium hydroxide: - The molar mass of barium hydroxide (Ba(OH)2) is 171.34 g/mol. - According to the balanced equation, 1 mole of sulfuric acid reacts with 1 mole of barium hydroxide. - Therefore, the number of moles of barium hydroxide is the same as the number of moles of sulfuric acid.

3. Calculate the mass of the precipitate (barium sulfate): - According to the balanced equation, the molar ratio between sulfuric acid and barium sulfate is 1:1. - The molar mass of barium sulfate (BaSO4) is 233.39 g/mol. - Therefore, the mass of the precipitate can be calculated as: mass of BaSO4 = moles of BaSO4 * molar mass of BaSO4

Now, let's calculate the mass of the precipitate using the given information.

Calculation:

1. Moles of sulfuric acid: - Moles of H2SO4 = 17 g / 98 g/mol - Moles of H2SO4 = 0.173 moles (approximately)

2. Moles of barium hydroxide: - Moles of Ba(OH)2 = Moles of H2SO4 (according to the balanced equation) - Moles of Ba(OH)2 = 0.173 moles (approximately)

3. Mass of the precipitate (barium sulfate): - Mass of BaSO4 = Moles of BaSO4 * Molar mass of BaSO4 - Mass of BaSO4 = 0.173 moles * 233.39 g/mol - Mass of BaSO4 = 40.36 grams (approximately)

Therefore, the mass of the precipitate (barium sulfate) formed in the reaction between 17 grams of sulfuric acid and barium hydroxide is approximately 40.36 grams.