Sodium azide (NaN3) yields N2 gas when heated to 300 degrees Celsius , a reaction used in automobile airbags. If 1.00 mol of N2 has a volume of 47.0 liters under the reaction conditions , how many liters of gas can be formed by heating 35g of NaN3?
The reaction is 2Nan3------>3N2(g)+2Na
35g of NaN3*(1 mole/65.01g of NaN3)= moles of NaN3
moles of NaN3*(3 moles of N2/2 moles of NaN3)=moles of N2
moles of N2*(47 L/1 mole of N2)= volume in L.
To solve this problem, we first need to find the molar mass of NaN3 (sodium azide).
The molar mass of Na is 22.99 g/mol.
The molar mass of N is 14.01 g/mol.
So, the molar mass of NaN3 is:
(22.99 g/mol) + 3(14.01 g/mol) = 65.00 g/mol.
Next, we need to find the number of moles of NaN3 in 35g.
Number of moles = Mass / Molar mass
Number of moles of NaN3 = 35g / 65.00 g/mol = 0.538 mol.
According to the balanced chemical equation, 2 mol of NaN3 yields 3 mol of N2 gas.
So, 0.538 mol of NaN3 will yield (3/2) * 0.538 mol of N2 gas = 0.807 mol of N2 gas.
Finally, we can use the ideal gas law to find the volume of N2 gas formed.
PV = nRT
V = nRT / P
where:
P = pressure (assuming constant and not given in the problem)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (300 degrees Celsius = 573 K)
V = (0.807 mol)(0.0821 L·atm/(mol·K))(573 K) / P
We are given that 1.00 mol of N2 gas has a volume of 47.0 liters. So, we can set up the following proportion:
1.00 mol / 47.0 L = 0.807 mol / V
Cross-multiplying and solving for V:
V = (0.807 mol * 47.0 L) / 1.00 mol
V = 37.929 L
Therefore, 35g of NaN3 will yield approximately 37.929 liters of N2 gas when heated to 300 degrees Celsius.
To solve this problem, we need to use stoichiometry to determine the number of moles of N2 gas produced when 35 grams of NaN3 reacts.
Step 1: Convert the mass of NaN3 to moles.
The molar mass of NaN3 is calculated by adding the molar masses of sodium (Na) and nitrogen (N) multiplied by their respective numbers in the formula.
Molar mass of NaN3 = (Na: 22.99 g/mol) + (N: 14.01 g/mol x 3) = 65.00 g/mol
To convert grams to moles, we use the formula:
moles = mass / molar mass
moles of NaN3 = 35 g / 65.00 g/mol ≈ 0.538 moles
Step 2: Use the balanced equation to determine the stoichiometry.
The balanced equation tells us that 2 moles of NaN3 react to produce 3 moles of N2 gas.
Therefore, the stoichiometry ratio is:
2 moles NaN3 : 3 moles N2
Step 3: Calculate the number of moles of N2 produced.
We can use the stoichiometry ratio to find the number of moles of N2 gas produced from 0.538 moles of NaN3.
moles of N2 = 0.538 moles NaN3 x (3 moles N2 / 2 moles NaN3) ≈ 0.807 moles N2
Step 4: Convert the number of moles of N2 to liters.
Under the reaction conditions given, 1.00 mol of N2 has a volume of 47.0 liters.
To convert moles to liters, we use the formula:
volume = moles x molar volume
volume of N2 = 0.807 moles x 47.0 L/mol ≈ 37.93 liters
Therefore, when 35 grams of NaN3 is heated, approximately 37.93 liters of N2 gas can be formed.