A mixture containing 9 moles of fluorine and 4 moles of sulfur is allowed to react to form sulfur hexafluoride. How many moles of fluorine remaines after 3 moles of sulfur reacted?

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The chemical formula of sulfur hexaflouride is `SF_6` which means for every mole of sulfur that becomes sulfur hexafouride, 6 moles of flourine must also react.  The ratio of S:F = 1:6, means that 3 moles of sulfur cannot react. Since there were only 9 moles of flourine to begin with, flourine is the limiting reactant.  Once those 9 moles have bonded with the sulfur, the reaction stops. 

n:9 = 1:6 can be rewritten in fractions `n/9 = 1/6` We cross-multiply and we get `6n=9` we divide each side by 6 `n=9/6` which simplifies to `n=1 1/2` So only 1 1/2 moles of sulfur can react with 9 moles of flourine to form sulfur hexaflouride. 

For 3 moles of sulfur to react we can use the ratios to find 3:f=1:6.  We re-write as fractions`(3/f) = 1/6` Cross multiply to get f=18.  At least 18 moles of flourine must be present for 3 moles of sulfur to react. 

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