Final answer:
In the reaction of 6 grams of lithium with 4 grams of sodium iodide, sodium iodide is the limiting reactant. No limiting reactant remains after the reaction as all 4 grams of sodium iodide will be completely consumed in the reaction.
Step-by-step explanation:
When 6 grams of lithium reacts with 4 grams of sodium iodide in a single displacement reaction, we need to determine the limiting reactant. First, we need the balanced reaction equation, which is 2Li + 2NaI → 2Na + 2LiI. To find which reactant is limiting, we convert grams to moles using their molar masses. Lithium has a molar mass of ~6.94 g/mol and sodium iodide has a molar mass of ~149.89 g/mol.
Number of moles of Li = 6 g ÷ 6.94 g/mol ≈ 0.8645 moles
Number of moles of NaI = 4 g ÷ 149.89 g/mol ≈ 0.0267 moles
According to the balanced equation, the reaction requires the mole ratio of Li to NaI to be 1:1. However, we have more moles of Li than NaI. Therefore, NaI is the limiting reactant. Since both reactants are consumed in equimolar amounts and, in this case, the number of moles of NaI is less, all of the NaI will be consumed with excess Li remaining.
Since all of the NaI is used up, there will be no limiting reactant remaining. Excess Li will be left over after the reaction. To know how much excess Li would be left, we would subtract the moles of Li that react with NaI from the initial moles of Li.