How many grams of lithium are required to completely react with 61.9 mL of N2 gas at STP
To determine the number of grams of lithium required to react with 61.9 mL of N2 gas, we need to use the balanced chemical equation for the reaction between lithium and nitrogen gas.
The balanced equation is:
6 Li + N2 → 2 Li3N
From the equation, we can see that 6 moles of lithium (6 Li) react with 1 mole of nitrogen gas (N2) to produce 2 moles of lithium nitride (Li3N).
First, let's convert the volume of nitrogen gas at STP (Standard Temperature and Pressure) to moles by using the ideal gas law:
PV = nRT,
where:
P = pressure = 1 atm,
V = volume = 61.9 mL = 0.0619 L,
n = moles of gas (N2),
R = ideal gas constant = 0.0821 L·atm/(mol·K),
T = temperature = 273 K (STP).
Rearranging the equation: n = PV/RT.
n = (1 atm) × (0.0619 L) / (0.0821 L·atm/(mol·K) × 273 K)
n = 0.002308 moles of N2 gas.
Now, we need to find the moles of lithium required for the reaction using stoichiometry.
From the balanced equation, we know that 6 moles of lithium (6 Li) react with 1 mole of nitrogen gas (N2).
Therefore, the moles of lithium required are:
moles of Li = (moles of N2) × (6 moles of Li / 1 mole of N2)
moles of Li = 0.002308 moles of N2 × (6 moles of Li / 1 mole of N2)
moles of Li = 0.01385 moles of Li.
Finally, to find the grams of lithium, we need to multiply the moles of lithium by the molar mass of lithium.
The molar mass of lithium is approximately 6.94 g/mol.
grams of Li = (moles of Li) × (molar mass of Li)
grams of Li = 0.01385 moles of Li × 6.94 g/mol
grams of Li = 0.09607 grams of Li (rounded to five decimal places).
Therefore, approximately 0.09607 grams of lithium are required to completely react with 61.9 mL of N2 gas at STP.
To find out how many grams of lithium are required to react with a certain volume of N2 gas at STP (Standard Temperature and Pressure), we need to use the concept of stoichiometry and the ideal gas law.
Step 1: Write the balanced equation for the reaction between lithium and nitrogen gas:
6Li + N2 -> 2Li3N
Step 2: Use the ideal gas law to find the number of moles of N2 gas:
PV = nRT
At STP (Standard Temperature and Pressure), the values are:
P = 1 atm
V = 61.9 mL = 61.9 / 1000 L = 0.0619 L
R = 0.0821 L·atm/(mol·K) (gas constant)
T = 273.15 K
Plugging in these values, we can solve for n (number of moles of N2 gas):
(1 atm) * (0.0619 L) = n * (0.0821 L·atm/(mol·K)) * (273.15 K)
n = (0.0619 L * 1 atm) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.002449 mol
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of lithium required. According to the balanced equation, it takes 3 moles of lithium to react with 1 mole of N2 gas.
So, 0.002449 mol N2 gas x (3 mol Li / 1 mol N2) = 0.007347 mol Li.
Step 4: Convert moles of lithium to grams using the molar mass of lithium. The molar mass of lithium is 6.94 g/mol.
So, 0.007347 mol Li x (6.94 g Li / 1 mol Li) = 0.051 g Li.
Therefore, approximately 0.051 grams of lithium are required to completely react with 61.9 mL of N2 gas at STP.