Upon decomposition, one sample of magnesium fluoride produced 1.65kg of magnesium and 2.58kg of fluorine. A second sample produce 1.38kg of magnesium. How much fluorine (in grams) did the second sample produce?
The ratio of Mg to F is constant.
f2 / m2 = f1 / m1
To find the second mass of fluorine (f2),
rearrange and substitute appropriate values.
You need to create an identity to solve for the mass of fluorine:
1.65 kg of Mg=2.58 kg of F
and
1.38 kg of Mg=x kg of F
So,
x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg
Solve for x kg of F,
x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg
x kg of F=2.157Kg of F
2.157 kg of F*(10^3g/1kg )=2.16 x 10^3 g of F or 2,160 g of F
Answer contains 3 sig figs.
1.98
2020
To find out how much fluorine the second sample produced, we need to use the information given about the decomposition of the first sample.
Let's assume that the first sample of magnesium fluoride (MgF2) decomposed completely according to the chemical equation:
2MgF2 -> 2Mg + F2
From the given information, we know that the first sample produced 1.65 kg of magnesium and 2.58 kg of fluorine. This means that for every 1.65 kg of magnesium produced, 2.58 kg of fluorine was produced.
To calculate the amount of fluorine produced by the second sample (let's call it MgF2'), we can set up a proportion using the ratio of magnesium to fluorine from the first sample:
1.65 kg Mg / 2.58 kg F = 1.38 kg Mg / x kg F
Cross-multiplying, we get:
1.65 kg Mg * x kg F = 2.58 kg F * 1.38 kg Mg
Simplifying:
x kg F = (2.58 kg F * 1.38 kg Mg) / 1.65 kg Mg
x kg F ≈ 2.15 kg F
Since we're looking for the amount of fluorine produced by the second sample in grams, we need to convert kilograms to grams by multiplying by 1000:
2.15 kg F * 1000 g/kg = 2150 g F
Therefore, the second sample of magnesium produced approximately 2150 grams of fluorine.
F/Mg = 2.58/1.65
So, given 1.28kg Mg, we should have
1.28*(2.58/1.65) = 2.00kg F