A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2Li + F₂$\to$ 2LiF.After the reaction is complete,what will be present? A)2.16 moles lithium fluoride only B)0.789 moles lithium fluoride only C)2.16 moles lithium fluoride and 0.395 moles fluorine D)0.789 moles lithium fluoride and 1.37 moles lithium E)none of these

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To solve this, first, determine the moles of lithium and fluorine. The molar mass of lithium (Li) is approximately 6.94 g/mol, and for fluorine (F2), it is approximately 38 g/mol (since fluorine is diatomic, its atomic mass is about 19 g/mol, so for F2, it's 19*2). For 15 g of Li: $ \frac{15 \, \text{g}}{6.94 \, \text{g/mol}} \approx 2.16 \, \text{moles of Li} $. For 15 g of F2: $ \frac{15 \, \text{g}}{38 \, \text{g/mol}} \approx 0.395 \, \text{moles of F2} $. The reaction shows that 2 moles of Li react with 1 mole of F2 to produce 2 moles of LiF. Since there are more moles of Li (2.16 moles) than needed to react with the available F2 (0.395 moles), Li is in excess, and F2 is the limiting reactant. All of the F2 will react, and $ 0.395 \, \text{moles of F2} \times 2 = 0.789 \, \text{moles of LiF} $ will be produced. The remaining Li moles can be calculated by subtracting the moles of Li that reacted (0.395 moles of F2 react with twice as many moles of Li, so $ 0.395 \times 2 = 0.79 \, \text{moles of Li} $ reacted) from the initial moles of Li, leaving $ 2.16 - 0.79 = 1.37 \, \text{moles of Li} $ unreacted.

A 15-G Sample of Lithium Is Reacted with 15 G →\to→

A 15-G Sample of Lithium Is Reacted with 15 G $\to$