Lithium and fluorine always react so that 7 g of lithium and 19.2 g of fluorine produce 26.2 g of lithium fluoride.
A)what mass of lithium floride forms when 1 g of lithium is added to 4 g of fluorine?
b)Determine the mass and the identity of the reactant remaining
The chemical reaction is 2Li + F2 = 2LiF The mass of lithium becomes 7 times smaller, then the mass of fluorine and lithium fluoride also becomes 7 times smaller. If 7 g of lithium reacts with 19.2 g of fluorine, then 1 g of lithium has to react with 2.74g (19.2 / 7). But we have 4 g of fluorine. It’s more than we need to complete the reaction. That’s why the limiting reactant is Li.
The mass of lithium fluoride is m(LiF) = 26.2 / 7 = 3.74 g
The mass of fluorine remaining: m(F2) = 4.0 - 2.74 = 1.26 g
Answer: 3.74 g of LiF forms in the reaction 1.26 g of F2 is remaining