Upon decomposition, one sample of magnesium fluoride produced 1.66kg of magnesium and 2.56kg of fluorine. A second sample produced 1.29kg of magnesium.
How much fluorine (in grams) did the second sample produce?
MgF2 ===> Mg + F2
..........1.66..2.56
..........1.29....x
(1.66/1.29) = (2.56/x)
Solve for x and convert to grams.
To find out how much fluorine the second sample produced, we need to use the stoichiometry of the reaction.
The chemical formula for magnesium fluoride is MgF2, which means each mole of magnesium fluoride produces one mole of magnesium (Mg) and two moles of fluorine (F2).
First, we need to calculate the number of moles of magnesium produced in the second sample. We know that the molar mass of magnesium is 24.31 g/mol. Therefore, we have:
1.29 kg x (1000 g/1 kg) = 1290 g of magnesium
1290 g / 24.31 g/mol = 53.09 mol of magnesium
Since the molar ratio between magnesium and fluorine is 1:2, we can conclude that we have 53.09 mol of magnesium, which means we have twice that number of moles of fluorine.
Therefore, the number of moles of fluorine produced in the second sample is 2 x 53.09 mol = 106.18 mol.
Now, we need to convert the moles of fluorine to grams. The molar mass of fluorine is 19.00 g/mol.
106.18 mol x 19.00 g/mol = 2017.42 g
Therefore, the second sample of magnesium produced 2017.42 grams of fluorine.