Respuesta :
The question is incomplete, here is the complete question:
[tex]Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)[/tex]
Xianming runs a titration and collects, dries, and weighs the [tex]BaSO_4(s)[/tex] produced in the experiment. He reports a mass of 0.2989 g go [tex]BaSO_4[/tex] Based on this, calculate the concentration of [tex]Ba(OH)_2[/tex] solution.
Answer: The concentration of barium hydroxide solution is 0.0013 moles.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium sulfate = 0.2989 g
Molar mass of barium sulfate = 47.87 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of barium sulfate}=\frac{0.2989g}{233.4g/mol}=0.0013mol[/tex]
The given chemical reaction follows:
[tex]Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)[/tex]
By Stoichiometry of the reaction:
1 mole of barium sulfate is produced from 1 mole of barium hydroxide
So, 0.0013 moles of barium sulfate will be produced from = [tex]\frac{1}{1}\times 0.0013=0.0013mol[/tex] of barium hydroxide
As, no volume of the container is given. So, the concentration will be calculated in moles only.
Hence, the concentration of barium hydroxide solution is 0.0013 moles.
