Question

If a solution of sodium sulfate is mixed with a solution of barium nitrate, what is the molar mass of the precipitate that is produced?

Answer 1

The molar mass of the precipitate is 233.39 g/mol.

Step-by-step explanation:

We are given that a solution of sodium sulfate (Na₂SO₄(aq)) is mixed with a solution of barium nitrate (Ba(NO₃)₂(aq)). And we want to determine the molar mass of the precipitate this is formed.

Write the balanced double-replacement reaction. Note that all sodium and nitrate ions are soluble and BaSO₄ is insoluble. Hence:

`\displaystyle \text{Na$_2$SO$_4$}_\text{(aq)} + \text{Ba(NO$_3$)$_2$}_\text{(aq)} \longrightarrow 2\text{NaNO$_3$}_\text{(aq)} + \text{BaSO$_4$}_\text{(s)}`

From the equation, we can see that the precipitate that is formed is barium sulfate.

The molar mass of barium sulfate is:

`\displaystyle \begin{aligned} \text{MW}_\text{BaSO$_4$} & = (137.33 + 32.06 + 4(16.00)) \text{ g/mol} \\ \\ & =233.39\text{ g/mol} \end{aligned}`

In conclusion, the molar mass of the precipitate is 233.39 g/mol.