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

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey