Question

How do you convert 64.3 g `PbBr_2` to moles?

Answer 1

`64.3` `"g"` of `PbBr_"2"` is `0.175` `"mols"`.

Explanation:

So the formula we are using is `n=m/M`.
=> Where `n` is the amount in mols.
=> Where `m` is the mass in grams.
=> Where `M` is the molar mass in `g/("mols")`.

First off, we need to find the molar mass of `PbBr_"2"`.

Using a Periodic Table, lead has a molar mass of `207.20 g/("mol")` and bromine has a molar mass of `79.90 g/("mol")`. Including the two atoms of bromine, when we add them all up, we get `367.00 g/("mol")`.

Now, we can plug them into the formula.

`n=m/M`

`n=64.3/367`

`n=0.175204359`

`n=0.175`

Therefore, `64.3` `"g"` of `PbBr_"2"` is `0.175` `"mols"`.

Hope this helps :)