When making a chemical; 50 g of Sulfur Hexafluoride (SFl6) are produced when 43 g of Sulfur is exposed to Fluorine. How many grams of Flourine are needed for this reaction?
To find out how many grams of fluorine are needed for this reaction, we need to determine the molar ratio between sulfur and sulfur hexafluoride.
From the balanced chemical equation:
S + 3F2 → SF6
We can see that for every 1 mole of sulfur (32 g), we need 3 moles of fluorine (19 g each) to produce 1 mole of sulfur hexafluoride (146 g).
Now, we can set up a proportion to find the amount of fluorine needed:
(3 moles of fluorine / 1 mole of sulfur) = (x grams of fluorine / 43 grams of sulfur)
Cross multiplying:
3 moles of fluorine * 43 grams of sulfur = x grams of fluorine * 1 mole of sulfur
x grams of fluorine = (3 moles of fluorine * 43 grams of sulfur) / 1 mole of sulfur
x grams of fluorine = 129 grams of sulfur / 1 mole of sulfur
x grams of fluorine = 129 grams of fluorine.
Therefore, 129 grams of fluorine are needed for this reaction.