Please help again. How many grams of lithium are required to completely react with 70.7 mL of N2 gas at STP when given the following reaction:
6Li + N2 = 2Li3N
Here is a worked example that will take care of all of your stoichiometry problems. Remember that 1 mole of gas at STP occupies 22.4L
http://www.jiskha.com/science/chemistry/stoichiometry.html
Thank you, Dr. Bob! Wish YOU would have written my chemistry book!
To determine the number of grams of lithium required to completely react with 70.7 mL of N2 gas at STP, you will need to follow these steps:
Step 1: Convert the volume of N2 gas to moles
You need to convert the volume of N2 gas to moles using the ideal gas law. At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. So, you can use this relationship to convert the volume to moles:
70.7 mL of N2 gas × (1 L / 1000 mL) × (1 mol / 22.4 L) = X mol
Step 2: Determine the stoichiometry of the reaction
From the given reaction: 6Li + N2 = 2Li3N, the stoichiometric ratio between lithium and nitrogen gas is 6:1. This means that for every 6 moles of lithium, you need 1 mole of N2 gas.
Step 3: Convert moles of N2 gas to moles of lithium
Since you now have the number of moles of N2 gas, you can use the stoichiometric ratio from step 2 to determine the number of moles of lithium required:
X mol N2 × (6 mol Li / 1 mol N2) = Y mol Li
Step 4: Convert moles of lithium to grams
To determine the mass of lithium required, you need to convert the number of moles to grams using the molar mass of lithium. The molar mass of lithium (Li) is approximately 6.94 g/mol.
Y mol Li × (6.94 g Li / 1 mol Li) = Z g Li
Therefore, Z grams of lithium are required to completely react with 70.7 mL of N2 gas at STP.
To determine the number of grams of lithium required to react with 70.7 mL of N2 gas at STP, we need to use stoichiometry.
First, let's convert the volume of N2 gas from mL to moles using the Ideal Gas Law. We can use the equation:
PV = nRT
Where:
P = pressure (STP conditions: 1 atm)
V = volume of gas (70.7 mL)
n = moles of gas (to be determined)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP conditions: 273.15 K)
Rearranging the equation, we get:
n = PV / RT
Substituting the values:
n = (1 atm) * (70.7 mL) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = 2.889 moles
Now we can use the balanced equation to establish the stoichiometric relationship between lithium (Li) and nitrogen gas (N2) in order to find the number of moles of lithium required.
From the balanced equation:
6Li + N2 = 2Li3N
We can see that 1 mole of N2 reacts with 6 moles of Li. Therefore, we can set up a ratio to calculate the number of moles of Li:
1 mole N2 : 6 moles Li
Using this ratio, we can find the moles of Li required:
moles Li = 2.889 moles N2 * (6 moles Li / 1 mole N2)
moles Li = 17.334 moles
Finally, convert moles of Li to grams by multiplying it by the molar mass of lithium (Li), which is approximately 6.94 g/mol:
grams Li = 17.334 moles Li * 6.94 g/mol
grams Li = 120.181 g
Therefore, approximately 120.181 grams of lithium are required to completely react with 70.7 mL of N2 gas at STP.