99.1k views
5 votes
how many grams of barium sulfate will be formed upon the complete reaction of 25.6 grams of barium hydroxide with excess sulfuric acid?

User Davekr
by
6.0k points

1 Answer

1 vote

Answer:

Mass of barium sulfate produced = 35.007 g

Step-by-step explanation:

Given data:

mass of Ba(OH)₂ = 25.6 g

mass of barium sulfate = ?

Solution:

First of all we will write the balance chemical equation:

Ba(OH)₂ + H₂SO₄ → BaSO₄ + 2H₂O

Number of moles of Ba(OH)₂ = mass/molar mass

Number of moles of Ba(OH)₂ = 25.6 g / 171.34 g/mol

Number of moles of Ba(OH)₂ = 0.15 mol

Now we will compare the moles of barium sulfate with barium hydroxide.

Ba(OH)₂ : BaSO₄

1 : 1

0.15 : 0.15

Mass of Barium sulfate:

Mass = number of moles × molar mass

Mass = 0.15 mol × 233.38 g/mol

Mass = 35.007 g

User Yathi
by
6.9k points