Respuesta :
This is an incomplete question, here is a complete question.
For the following reaction, 3.81 grams of sulfuric acid are mixed with excess barium hydroxide. The reaction yields 8.45 grams of barium sulfate.
[tex]Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O[/tex]
What the is theoretical mass of barium sulfate?
Answer: The theoretical mass of [tex]BaSO_4[/tex] is, 9.08 grams.
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 3.81 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
First we have to calculate the moles of [tex]H_2SO_4[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{3.81g}{98g/mol}=0.0389mol[/tex]
Now we have to calculate the moles of [tex]BaSO_4[/tex]
The balanced chemical equation is:
[tex]Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 1 mole of [tex]BaSO_4[/tex]
So, 0.0389 mole of [tex]H_2SO_4[/tex] react to give 0.0389 mole of [tex]BaSO_4[/tex]
Now we have to calculate the mass of [tex]BaSO_4[/tex]
[tex]\text{ Mass of }BaSO_4=\text{ Moles of }BaSO_4\times \text{ Molar mass of }BaSO_4[/tex]
Molar mass of [tex]BaSO_4[/tex] = 233.38 g/mole
[tex]\text{ Mass of }BaSO_4=(0.0389moles)\times (233.38g/mole)=9.08g[/tex]
Therefore, the theoretical mass of [tex]BaSO_4[/tex] is, 9.08 grams.
