Upon decomposition, one sample of Magnesium Fluoride produced 1.65 kg of magnesium and 2.57 kg of fluorine. A second sample produces 1.33 kg of magnesium. How much fluorine (in grams) did the second sample produce?
......MgF2 ==> Mg + F2
..............1.65..2.57
Therefore, MgF2 must have started at 1.65 + 2.57 = 4.22 kg.
Second sample produced 1.33 kg Mg.
F2 produced = 2/57 x (1.33/1.65) = ? kg. Convert to grams.
2350
To find the amount of fluorine produced by the second sample, we can use the ratio of magnesium to fluorine obtained from the first sample.
Let's start by calculating the ratio of magnesium to fluorine from the first sample:
1. First, note that the atomic mass of magnesium (Mg) is 24.31 g/mol, and the atomic mass of fluorine (F) is 18.99 g/mol.
2. Calculate the molar mass of magnesium fluoride (MgF₂) by adding the atomic masses of magnesium and two fluorine atoms:
Molar mass of MgF₂ = (24.31 g/mol) + 2 * (18.99 g/mol) = 62.29 g/mol.
3. Use the molar mass of magnesium fluoride to convert the masses of magnesium and fluorine produced by the first sample into moles:
Moles of magnesium from first sample = 1650 g / 24.31 g/mol = 67.79 mol.
Moles of fluorine from first sample = 2570 g / 18.99 g/mol = 135.6 mol.
4. Calculate the ratio of moles between magnesium and fluorine from the first sample:
Ratio (Mg:F) from first sample = Moles of magnesium / Moles of fluorine
= 67.79 mol / 135.6 mol
≈ 0.4999 (rounded to four decimal places).
Now, we can use the obtained ratio to find the amount of fluorine produced by the second sample:
5. Convert the mass of magnesium produced by the second sample into moles using the molar mass of magnesium:
Moles of magnesium from second sample = 1330 g / 24.31 g/mol
≈ 54.73 mol.
6. Multiply the moles of magnesium from the second sample by the ratio (Mg:F) calculated from the first sample to find the moles of fluorine:
Moles of fluorine from second sample = Moles of magnesium from second sample * (Mg:F) from first sample
= 54.73 mol * 0.4999
≈ 27.36 mol.
7. Finally, convert the moles of fluorine from the second sample back into grams by multiplying by the molar mass of fluorine:
Mass of fluorine from second sample = Moles of fluorine from second sample * 18.99 g/mol
≈ 27.36 mol * 18.99 g/mol
≈ 519.67 g.
Therefore, the second sample of magnesium fluoride produced approximately 519.67 grams of fluorine.