Upon decomposition, one sample of magnesium fluoride produced 0.764 kg of magnesium and 1.19 kg of fluorine. A second sample has a total mass of 5.75 kg. How much magnesium did the second sample produce?

w Mg = 2.248 Kg

Explanation:

MgF2 → Mg + 2F

∴ mass Mg = 0.764 Kg

∴ mass F = 1.19 Kg

⇒ %wt Mg = (0.764 / (0.764+1.19)) * 100 = 39.099%

if w sample = 5.75 Kg

⇒ 39.099 % = (w Mg/5.75) * 100

⇒ 0.39099 = w Mg/5.75 Kg

⇒ (5.75 Kg) (0.39099) = w Mg

⇒ w Mg = 2.248 Kg

3.50 kg

Explanation:

Law of definite proportion: These law states that all pure samples of a particular chemical compound contains similar elements combined in the same proportion by mass.

From the law above,

First sample,

The ratio of fluorine and magnesium in the sample of magnesium fluoride is

0.764:1.19 = 1:1.6

mass of fluorine: mass of magnesium ≈ 1:1.6

Second sample,

mass of magnesium fluoride = 5.75 kg

mass of magnesium = 1.19 (5.75) / (1.19+0.764)

mass of magnesium = 6.8425/1.954

mass of magnesium = 3.50 kg

Hence the second sample produced 3.50 kg of magnesium.