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?

Answer 1

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.