Lithium reacts with nitrogen gas according to the following reaction:
6Li(s)+N2(g)→2Li3N(s)
What mass of lithium (in g) is required to react completely with 59.4 mL of N2 gas at STP?
6Li(s)+N2(g)→2Li3N(s)
mols N2 = 59.6 x *1 mol/22,400 L) = approx 0.0027 but that's an estimate.
mols Li = mols N2 x (6 mols Li/1 mol N2) = approx 0.0027 x 6 = ?
Then g Li = mols Li x atomic mass Li = ?
Post your work if you get stuck.
To determine the mass of lithium required, we can use the balanced chemical equation and the ideal gas law.
Step 1: Convert volume to moles of nitrogen gas
Given volume of N2 gas is 59.4 mL.
We need to convert mL to L, so divide by 1000:
59.4 mL ÷ 1000 = 0.0594 L
Next, we can use the ideal gas law equation:
PV = nRT
At STP (Standard Temperature and Pressure), the values are:
P = 1 atm
V = 0.0594 L
R = 0.0821 L·atm/(mol·K)
T = 273.15 K (0 degrees Celsius)
Plugging in the values:
(1 atm)(0.0594 L) = n (0.0821 L·atm/(mol·K))(273.15 K)
Solving for n (moles of N2 gas):
n = (1 atm × 0.0594 L) / (0.0821 L·atm/(mol·K) × 273.15 K)
n ≈ 0.0022 mol
Step 2: Calculate the moles of lithium required
According to the balanced equation, 6 moles of Li reacts with 1 mole of N2 gas.
Therefore, the moles of Li required will be:
0.0022 mol of N2 × (6 mol of Li / 1 mol of N2) = 0.0132 mol of Li
Step 3: Convert moles to grams
To convert moles to grams, we need to use the molar mass of lithium (Li), which is approximately 6.941 g/mol.
0.0132 mol × 6.941 g/mol = 0.0917 g
Therefore, approximately 0.0917 grams of lithium is required to react completely with 59.4 mL of N2 gas at STP.
To find the mass of lithium required to react completely with nitrogen gas, we first need to determine the number of moles of nitrogen gas using the given volume at STP.
STP (Standard Temperature and Pressure) conditions are defined as 0°C (273.15 K) and 1 atm of pressure.
Step 1: Convert the given volume of N2 gas to moles.
At STP, one mole of any ideal gas occupies 22.4 Liters. Therefore, we need to convert the volume of nitrogen gas from milliliters (mL) to liters (L) and then to moles.
Given volume of N2 gas = 59.4 mL
Converting mL to L:
1 L = 1000 mL
59.4 mL = 59.4/1000 L = 0.0594 L
Converting volume to moles:
1 mole of any ideal gas = 22.4 L
moles of N2 gas = volume (L) / 22.4 L/mol
moles of N2 gas = 0.0594 L / 22.4 L/mol
Step 2: Calculate the moles of lithium required using the stoichiometric coefficients of the balanced equation.
From the balanced equation, we can see that 1 mole of nitrogen gas reacts with 6 moles of lithium (6Li).
moles of lithium = moles of N2 gas * (6 moles of Li / 1 mole of N2 gas)
Finally, we can calculate the mass of lithium required using the molar mass of lithium (6Li) which is 6.941 g/mol.
mass of lithium (g) = moles of lithium * molar mass of lithium
Now, let's plug in the numbers and calculate:
moles of N2 gas = 0.0594 L / 22.4 L/mol
moles of lithium = (0.0594 L / 22.4 L/mol) * (6 moles of Li / 1 mole of N2 gas)
mass of lithium (g) = moles of lithium * molar mass of lithium
mass of lithium (g) = [(0.0594 L / 22.4 L/mol) * (6 moles of Li / 1 mole of N2 gas)] * 6.941 g/mol