Question

Xianming runs a titration and collects, dries, and weighs the BaSO4(s) produced in the experiment. He reports a mass of 0.2989 g go BaSO4. Based on this, calculate the concentration of Ba(OH)2 solution.

Answer 1

The question is incomplete, here is the complete question:

`Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)`

Xianming runs a titration and collects, dries, and weighs the
`BaSO_4(s)` produced in the experiment. He reports a mass of 0.2989 g go
`BaSO_4` Based on this, calculate the concentration of
`Ba(OH)_2` solution.

Answer: The concentration of barium hydroxide solution is 0.0013 moles.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of barium sulfate = 0.2989 g

Molar mass of barium sulfate = 47.87 g/mol

Putting values in above equation, we get:

`\text{Moles of barium sulfate}=(0.2989g)/(233.4g/mol)=0.0013mol`

The given chemical reaction follows:

`Ba(OH)_2(aq.)+H_2SO_4(aq.)\rightarrow BaSO_4(s)+2H_2O(l)`

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 =
`(1)/(1)* 0.0013=0.0013mol` 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.