What volume of nitrogen gas at STP can form from 1.20 mol NaN3?
2NaN3(s) 2Na(s) + 3N2(g)
A. 40.3 L
B. 1.20 L
C. 13.4 L
D. 1.80 L
The answer is 40.3
Ok so with 1.2 mols of NaN3 you have 1.8 mols of nitrogen gas. How do I convert it to liters?
R, universal gas constant is 8.314 J/ Kmole or 0.0821 L atm/K mol.
in this case, apply 0.0821 for R since we are using liter and atm.
If you calculate it right, it will give 40.3 L
Well, here's a nitrogen fact for you: Nitrogen gas makes up about 78% of Earth's atmosphere. So, it's no joke that nitrogen is all around us!
Now, let's do some calculations to find out the volume of nitrogen gas that can form from 1.20 mol of NaN3 at STP (standard temperature and pressure). According to the balanced equation, 2 moles of NaN3 will produce 3 moles of N2 gas.
So, if 2 moles of NaN3 produces 3 moles of N2 gas, then 1.20 moles of NaN3 will produce (1.20 mol * 3 mol of N2 / 2 mol of NaN3) = 1.80 moles of N2 gas.
At STP, 1 mole of any gas occupies 22.4 liters of space. Therefore, 1.80 moles of N2 gas will occupy (1.80 mol * 22.4 L/mol) = 40.3 liters of space.
So, the correct answer is A. 40.3 L.
To find the volume of nitrogen gas at STP (Standard Temperature and Pressure) formed from 1.20 mol of NaN3, you need to use the balanced chemical equation and stoichiometry.
The balanced chemical equation is:
2NaN3(s) 2Na(s) + 3N2(g)
From the equation, you can see that 2 moles of NaN3 produce 3 moles of N2 gas.
To find the moles of N2 gas produced, you need to use stoichiometry to convert the moles of NaN3 to moles of N2.
Given:
Moles of NaN3 = 1.20 mol
Using stoichiometry, you can set up a proportion:
2 mol NaN3 = 3 mol N2
1.20 mol NaN3 = x mol N2
Solving for x:
x = (1.20 mol NaN3) * (3 mol N2 / 2 mol NaN3)
x = 1.80 mol N2
So, 1.20 mol of NaN3 produces 1.80 mol of N2 gas.
Now, to find the volume of N2 gas at STP, you need to use the ideal gas law:
PV = nRT
Where:
P = Pressure (STP is defined as 1 atm)
V = Volume
n = moles of gas
R = Ideal Gas Constant (0.0821 L·atm/(mol·K))
T = Temperature (STP is defined as 273.15 K)
Plugging in the values:
P = 1 atm
n = 1.80 mol
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
Now, solve for V:
V = (nRT) / P
V = (1.80 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V = 39.98 L or approximately 40.0 L
Therefore, the volume of nitrogen gas at STP formed from 1.20 mol of NaN3 is approximately 40.0 L.
The answer is A. 40.3 L
You can convert any mol in this equation to any other mol in this equation by using the coefficients in the balanced equation.
For example:
How many mols NaN3 are necessary to obtain 10 mols Na?
10 mols Na x (2 mols NaN3/2 mols Na) = 10 x 2/2 = 10.
The number on the top of the fraction is ALWAYS the coefficient of what you want to convert TO and the number on the bottom of the fraction is ALWAYS the coefficient of what you HAVE.