Final answer:
To determine the mass of lithium needed to react with 56.8 mL of N₂ gas at STP, we calculate the moles of N₂ and use stoichiometry to find the moles of Li required, then convert that to mass. The correct value calculates to approximately 0.1057 g Li, which is not listed among the provided options, indicating a possible error in the question or options.
Step-by-step explanation:
To calculate the mass of lithium required to react completely with 56.8 mL of N₂ gas at STP, we can use the ideal gas law and stoichiometry. The reaction is as follows: 6Li(s) + N₂(g) → 2Li₃N(s).
First, we convert the volume of N₂ to moles at STP using the ideal gas law, where 1 mole of any gas occupies 22.4 L at STP.
Number of moles of N₂ = 56.8 mL × (1 L / 1000 mL) × (1 mol / 22.4 L) ≈ 0.00254 mol
Using the stoichiometry of the balanced equation, we determine the moles of Li needed:
0.00254 mol N₂ × (6 mol Li / 1 mol N₂) ≈ 0.01524 mol Li
Lastly, we convert moles of Li to grams using the molar mass of Li:
0.01524 mol Li × (6.94 g/mol Li) ≈ 0.1057 g Li
This value indicates that the correct answer would have been approximately 0.1057 g, but this value does not match any of the given options. There might be an error either in the question or options provided.