Lithium reacts with nitrogen gas according to the following reaction: 6Li(s)+N2(g)-->2LiN(s)
What mass of lithium (in g ) is required to react completely with 59.9 mL of N2 gas at STP?
.110
0.107
Why did the lithium join a band? Because it wanted to be a Nitrogen Rockstar! Now let's calculate the mass required to become a chemical rockstar.
First, let's convert the volume of nitrogen gas into moles. At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. Since we have 59.9 mL of nitrogen gas, we can divide it by 1000 to convert it into liters: 59.9 mL = 0.0599 L.
Using the ideal gas law, PV = nRT, where P is the pressure (STP is 1 atm), V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature (273 K at STP), we can solve for n:
n = (P * V) / (R * T)
= (1 atm * 0.0599 L) / (0.0821 L·atm/(mol·K) * 273 K)
≈ 0.0278 moles
According to the balanced equation, 6 moles of lithium react with 1 mole of nitrogen gas to form 2 moles of lithium nitride. So, using stoichiometry, we can convert moles of nitrogen gas to moles of lithium:
0.0278 moles N2 * (6 moles Li / 1 mole N2) = 0.1667 moles Li
Finally, we need to convert moles of lithium to grams. The molar mass of lithium is 6.94 g/mol:
0.1667 moles Li * 6.94 g/mol ≈ 1.16 grams
Therefore, approximately 1.16 grams of lithium is required to react completely with 59.9 mL of nitrogen gas at STP. Rock on, lithium!
To find the mass of lithium required to react completely with nitrogen gas, we need to follow these steps:
Step 1: Determine the number of moles of nitrogen gas (N2) using the ideal gas equation.
At STP (standard temperature and pressure):
Temperature (T) = 273.15 K
Pressure (P) = 1 atm
Volume (V) = 59.9 mL = 0.0599 L (convert mL to L)
Using the ideal gas equation, PV = nRT, where R is the ideal gas constant (0.0821 L·atm/mol·K), we can solve for n (moles).
PV = nRT
(1 atm)(0.0599 L) = n(0.0821 L·atm/mol·K)(273.15 K)
n ≈ 0.00246 mol (rounded to 4 significant figures)
Step 2: Find the number of moles of lithium (Li) required using the balanced chemical equation.
From the balanced chemical equation: 6 moles of Li react with 1 mole of N2 to produce 2 moles of LiN.
So, using the stoichiometry, we can calculate the moles of Li required:
0.00246 mol N2 × (6 mol Li / 1 mol N2) ≈ 0.0148 mol Li (rounded to 3 significant figures)
Step 3: Calculate the mass of lithium required using the molar mass.
The molar mass of lithium (Li) is approximately 6.94 g/mol.
Mass of Li = moles of Li × molar mass of Li
Mass of Li = 0.0148 mol × 6.94 g/mol ≈ 0.1027 g (rounded to 4 significant figures)
Therefore, approximately 0.1027 grams of lithium is required to react completely with 59.9 mL of nitrogen gas at STP.
.555 g
Use the ideal gas law..
PV=nRT
and solve for moles of N2
SO....
(1.00atm)(.0599L)=n(0.0821L*atm/mol*K)(273K)
solve for n.
then use "n" moles of N2 and then set up an conversion factor.
so....
___moles of N2 x( 6mol Li/2mol N2) x (MolarMass Li/1mol of Li)
Then that gives you your mass of Lithium required to react.
Hope that helps!