Upon decomposition, one sample of magnesium fluoride produced 1.65kg of magnesium and 2.58kg of fluorine. A second sample produced 1.38kg of magnesium.
How much fluorine (in grams) did the second sample produce?
(1.65/2.58) = (1.38/x)
Solve for x.
Well, first of all, I have to say that magnesium and fluorine seem to be quite the couple in this decomposition. But let's crunch some numbers, shall we?
Since we know that the first sample produced 1.65kg of magnesium and 2.58kg of fluorine, we can use this ratio to find out how much fluorine the second sample produced.
The ratio of magnesium to fluorine in the first sample is 1.65kg/2.58kg. So, for every kilogram of magnesium, we have 2.58/1.65 kilograms of fluorine.
Now, in the second sample, we have 1.38kg of magnesium. To find out how much fluorine it produced, we can multiply the mass of magnesium by the ratio we calculated.
1.38kg of magnesium * (2.58kg/1.65kg) = 2.16kg of fluorine.
But wait, the question asks for the amount of fluorine in grams! So let's convert 2.16kg to grams.
2.16kg * 1000g/kg = 2160g.
So, the second sample produced 2160 grams of fluorine. Quite a mouthful, don't you think?
To find out how much fluorine the second sample produced, we need to use the information given in the question.
Let's start by determining the molar mass of magnesium fluoride. The molar mass of magnesium (Mg) is 24.31 g/mol and the molar mass of fluorine (F) is 18.99 g/mol. Since magnesium fluoride has one atom of magnesium and two atoms of fluorine, its molar mass is:
Molar mass of magnesium fluoride (MgF2) = (1 × molar mass of magnesium) + (2 × molar mass of fluorine)
= (1 × 24.31 g/mol) + (2 × 18.99 g/mol)
= 24.31 g/mol + 37.98 g/mol
= 62.29 g/mol
Next, we can use the molar mass of magnesium fluoride to convert the mass of magnesium produced in the second sample into moles. We divide the mass of magnesium (1.38 kg) by its molar mass (24.31 g/mol):
Moles of magnesium = mass of magnesium / molar mass of magnesium
= 1,380 g / 24.31 g/mol
= 56.77 mol
Since the stoichiometric ratio between magnesium and fluorine in magnesium fluoride is 1:2, for every 1 mole of magnesium, we should have 2 moles of fluorine.
Therefore, the number of moles of fluorine produced in the second sample is twice the number of moles of magnesium:
Moles of fluorine = 2 × moles of magnesium
= 2 × 56.77 mol
= 113.54 mol
Finally, we convert the moles of fluorine back into grams by multiplying by its molar mass:
Mass of fluorine = moles of fluorine × molar mass of fluorine
= 113.54 mol × 18.99 g/mol
= 2,155.04 g
Therefore, the second sample produced 2,155.04 grams of fluorine.
MgF2 produce
Sample 1) 1.65 kg Mg
2) 2.57 kg F
Sample 1) 1.32 kg Mg
2) ? g F
1. Find the ratio from sample 1,
1.65kg / 2,57kg = 0.64
2. Using Law of define properties, the first and second should have the same ratio.
1.3 kg / x kg F = 0.64
x = 2.0625 kg F
= 2062.5 g F