1.)If you fully react 23.grams of S. how much Li_(2)S would you produce from the reaction? 1S_(0)+16Liarrow 8Li_(2)S

To solve this problem, we need to use the balanced chemical equation and the given information to determine the amount of $Li_{2}S$ produced.<br /><br />Given information:<br />- The balanced chemical equation is: $1S_{0}+16Li\rightarrow 8Li_{2}S$<br />- The amount of $S$ reacted is 23 grams.<br /><br />Step 1: Calculate the molar mass of $S$.<br />The molar mass of $S$ is 32.06 g/mol.<br /><br />Step 2: Calculate the number of moles of $S$ reacted.<br />Number of moles of $S$ = Mass of $S$ / Molar mass of $S$<br />Number of moles of $S$ = 23 g / 32.06 g/mol = 0.714 mol<br /><br />Step 3: Use the balanced chemical equation to determine the number of moles of $Li_{2}S$ produced.<br />According to the balanced equation, 1 mole of $S$ produces 8 moles of $Li_{2}S$.<br />Number of moles of $Li_{2}S$ produced = 0.714 mol × 8 = 5.712 mol<br /><br />Step 4: Calculate the molar mass of $Li_{2}S$.<br />The molar mass of $Li_{2}S$ is 45.92 g/mol.<br /><br />Step 5: Calculate the mass of $Li_{2}S$ produced.<br />Mass of $Li_{2}S$ = Number of moles of $Li_{2}S$ × Molar mass of $Li_{2}S$<br />Mass of $Li_{2}S$ = 5.712 mol × 45.92 g/mol = 261.6 g<br /><br />Therefore, if you fully react 23 grams of $S$, you would produce 261.6 grams of $Li_{2}S$ from the reaction.