Question

1)Upon decomposition, one sample of magnesium fluoride produced 1.55 kg of magnesium and 2.42kg of fluorine. A second sample produced 1.00 kg of magnesium. How much fluorine (in grams) did the second sample produce?

Answer 1

Mass of magnesium = 1.55 kg =1550 g

Now

Molar Mass of magnesium = 24.305 g/mol

Moles of magnesium = Mass of magnesium/ Molar Mass of

magnesium

= 1550 g/24.305 g/mol

= 63.77 moles magnesium is produced

Mass of fluorine = 2.42 kg =2420 g

Now

Molar Mass of fluorine = 18.998 g/mol

Moles of fluorine = Mass of fluorine / Molar Mass of fluorine

= 2420 g /18.998 g/mol

= 127.38 moles fluorine is produced

Another sample

Mass of magnesium = 1.00 kg =1000 g

Moles of magnesium = Mass of magnesium/ Molar Mass of magnesium

= 1000 g /24.305 g/mol

= 41.14 moles magnesium is produced

Now

63.77 moles magnesium ==> 127.38 moles fluorine

41.14 moles magnesium ==>

41.14 moles magnesium * 127.38 moles fluorine / 63.77 moles magnesium

= 82.18 moles fluorine is produced

Mass of fluorine = Moles of fluorine * Molar Mass of fluorine

= 82.18 moles * 18.998 g/mol

= 1561 g fluorine is produced.