Question

What mass of metal is present in a `45.8*g` mass of barium sulfate?

Answer 1

There are approx. `27*g` of barium metal in such a mass of barium sulfate.

Explanation:

`"Moles of substance"` `=` `"Mass"/"Molar mass"`

`=(45.8*g)/(233.43*g*mol^-1)=0.0196*mol`.

Given that there are `0.0196*mol` of barium salt, we KNOW that there are also `0.0196*mol` of barium metal, and `4xx0.0196*mol` of oxygen atoms, and `0.0196*mol` of sulphur atoms. Are we clear on this? Because this is something that you should appreciate fundamentally, and (almost) without thinking.

And to find the mass of metal it is simply the product:

`"Moles of substance"xx"molar mass (of barium metal)"=0.0196*molxx137.33*g*mol^-1~=27*g`

The so-called barium meal you take before an X-ray is not entirely pleasant.