Question #eaa82

2 Answers
Dec 2, 2016

Answer:

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#

Dec 2, 2016

Answer:

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 #