Question

How much g of pure zinc will be obtained when 8.15g pure ZnO IS reduced by carbon reduction process.....?

Answer 1

`6.54 \ "g"` of zinc.

Explanation:

We have the following equation:

`ZnO(s)+C(s)->Zn(s)+CO(g)uparrow`

A blast furnace process, common in industries, to manufacture pure metals from their ores...

From here, we see that the mole ratio between zinc `(Zn)` and zinc oxide `(ZnO)` is `1:1=1`. So, one mole of zinc oxide produces one mole of pure zinc.

We have `8.15 \ "g"` of zinc oxide, so let's convert that amount into moles.

Zinc oxide has a molar mass of `81.38 \ "g/mol"`, so here, we get:

`(8.15color(red)cancelcolor(black)"g")/(81.38color(red)cancelcolor(black)"g""/mol")~~0.1 \ "mol"`

Since the mole ratio is `1`, then `0.1` moles of pure zinc is produced.

To convert that into grams, we multiply by zinc's molar mass, which is `65.38 \ "g/mol"`, and we produce:

`0.1color(red)cancelcolor(black)"mol"*(65.38 \ "g")/(color(red)cancelcolor(black)"mol")~~6.54 \ "g"`