Upon decomposition, one sample of magnesium fluoride produced 1.65kg of magnesium and 2.58kg of fluorine. A second sample produced 1.50 kg of magnesium. How much fluorine (in grams) did the second sample produce?
2.58 kg x (1.5/1.65) = ?kg F2
How many moles of water would be produced if 5 moles of propane were combusted?
To find out how much fluorine the second sample produced, we need to use the information from the first sample to determine the ratio of fluorine to magnesium in magnesium fluoride.
In the first sample, we have 1.65 kg of magnesium and 2.58 kg of fluorine. To determine the ratio, we divide the mass of fluorine by the mass of magnesium:
Fluorine to magnesium ratio = 2.58 kg / 1.65 kg = 1.56
This means that for every 1.56 kg of magnesium, we have 2.58 kg of fluorine in magnesium fluoride.
Now, let's use this ratio to find out how much fluorine the second sample produced.
Given that the second sample produced 1.50 kg of magnesium, we can set up a proportion using the ratio we determined earlier:
Fluorine produced / Magnesium produced = Fluorine to magnesium ratio
Let's solve for the amount of fluorine produced:
Fluorine produced = Magnesium produced * Fluorine to magnesium ratio
= 1.50 kg * 1.56
Fluorine produced = 2.34 kg
To convert 2.34 kg to grams, we multiply by 1000:
Fluorine produced in grams = 2.34 kg * 1000 = 2340 grams
Therefore, the second sample produced 2340 grams of fluorine.