What mass of lithium in grams is required to react completely with 57.9mL of N2 gas at STP?
6Li(s)+N2 (g)yields 2Li3N (s)
How many moles of N2 gas do you have. You need 6 times that number, to equal the number of moles of Lithium.
To determine the mass of lithium required to react completely with 57.9 mL of N2 gas at STP (Standard Temperature and Pressure), we need to follow a few steps:
Step 1: Balance the chemical equation:
The given balanced equation is: 6Li(s) + N2(g) → 2Li3N(s)
Step 2: Convert the given volume of N2 gas to moles:
Given that the volume of N2 gas is 57.9 mL, we need to convert it to moles. To do this, we use the Ideal Gas Law equation:
PV = nRT
P = pressure (STP has a pressure of 1 atm)
V = volume (57.9 mL)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP has a temperature of 273 K)
Rearranging the equation to solve for n:
n = PV / RT
n = (1 atm * 0.0579 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.0021 moles
Step 3: Determine the molar ratio between N2 and Li:
By examining the balanced chemical equation, we see that the ratio between N2 and Li is 1:6. This means that for every 1 mole of N2, we need 6 moles of Li.
Step 4: Calculate the moles of Li required:
Since the molar ratio between N2 and Li is 1:6 and we know that we have 0.0021 moles of N2, we can calculate the moles of Li required:
moles of Li = moles of N2 * (6 moles of Li / 1 mole of N2)
moles of Li = 0.0021 * 6
moles of Li ≈ 0.0126 moles
Step 5: Convert moles of Li to grams:
To convert moles of Li to grams, we need to use the molar mass of Li. The molar mass of Li is approximately 6.94 g/mol.
mass of Li = moles of Li * molar mass of Li
mass of Li = 0.0126 moles * 6.94 g/mol
mass of Li ≈ 0.0876 grams
Therefore, approximately 0.0876 grams of lithium are required to react completely with 57.9 mL of N2 gas at STP.