How many grams of barium sulfate will be formed upon the complete reaction of 25.6 grams of barium hydroxide with excess sulfuric acid?

Mass of barium sulfate produced = 35.007 g

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