Question

Question #eaa82

Answer 1

Approx. `4*g`

Explanation:

The percentage mass of magnesium in magnesium carbonate, `MgCO_3`:

`="Molar mass of metal"/"Molar mass of magnesium carbonate"=(24.305*g*mol^-1)/(84.32*g*mol^-1)xx100%=28.8%`

And thus in a `13.9*g` sample, there are `28.8%xx13.9*g~=4*g`

Answer 2

The Answer is approximately
`~= 4 g`

Explanation:

There is another way to consider this question Atomic wt of all atoms in the compound are as follows:
`C=12, Mg=24, O=16`
Wt of compound = `CMgO3 = 12 + 24 + 16*(3) = 84`
It means 84 gm CMgO3 gives 24 gm Magnesium
Therefore 13.9 g CMgO3 gives `24/84 *13.9 = 3.97 ~=4 gm`