utomotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements.
2 NaN3(s) → 2 Na(s) + 3 N2(g)
(a) How many moles of N2 are produced by the decomposition of 3.50 moles of NaN3?
(b) How many grams of NaN3 are required to form 16.0 g of nitrogen gas?
(c) How many grams of NaN3 are required to produce 19.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/L?
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To answer these questions, we will use the balanced equation for the decomposition of sodium azide as given:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
(a) To find the number of moles of N2 produced, we can use the mole ratio between NaN3 and N2. According to the balanced equation, 2 moles of NaN3 produce 3 moles of N2. Therefore, we can set up the following proportion:
2 moles of NaN3 / 3 moles of N2 = 3.50 moles of NaN3 / x moles of N2
Cross-multiplying and solving for x, we get:
x = (3.50 moles of NaN3) x (3 moles of N2 / 2 moles of NaN3)
x = 5.25 moles of N2
Therefore, 3.50 moles of NaN3 will produce 5.25 moles of N2.
(b) To find the number of grams of NaN3 required to form 16.0 g of nitrogen gas, we need to use the molar mass of NaN3 and the molar mass of N2. The molar mass of NaN3 is the sum of the atomic masses of sodium (Na) and nitrogen (N):
Molar mass of NaN3 = 22.99 g/mol (Na) + 3(14.01 g/mol) (N)
= 65.00 g/mol
Using the balanced equation, we know that 2 moles of NaN3 produce 3 moles of N2. Therefore, we can set up the following proportion:
2 moles of NaN3 / 3 moles of N2 = y g of NaN3 / 16.0 g of N2
Cross-multiplying and solving for y, we get:
y = (16.0 g of N2) x (2 moles of NaN3 / 3 moles of N2) x (65.00 g/mol)
y ≈ 69.33 g of NaN3
Therefore, approximately 69.33 grams of NaN3 are required to form 16.0 grams of nitrogen gas.
(c) To find the number of grams of NaN3 required to produce 19.0 ft3 of nitrogen gas with a density of 1.25 g/L, we can first convert the volume of nitrogen gas to liters. Since 1 ft3 is approximately 28.32 liters, we have:
19.0 ft3 * 28.32 L/ft3 ≈ 537.48 L
Next, we can calculate the number of moles of nitrogen gas using the ideal gas law equation:
PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we are given the volume (V), density (D), and molar mass (M) of nitrogen gas, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT = (D * M) / R
Given that the density of the gas is 1.25 g/L, the molar mass of N2 is 28.02 g/mol, and the ideal gas constant is 0.08206 L·atm/(mol·K), we can calculate n as:
n = (1.25 g/L * 537.48 L) / (0.08206 L·atm/(mol·K) * 298 K)
n ≈ 90.56 moles of N2
Since 2 moles of NaN3 produce 3 moles of N2, we can set up the following proportion:
2 moles of NaN3 / 3 moles of N2 = z g of NaN3 / 90.56 moles of N2
Cross-multiplying and solving for z, we get:
z = (90.56 moles of N2) x (2 moles of NaN3 / 3 moles of N2) x (65.00 g/mol)
z ≈ 119.79 g of NaN3
Therefore, approximately 119.79 grams of NaN3 are required to produce 19.0 ft3 of nitrogen gas with a density of 1.25 g/L.
(a) To determine the number of moles of N2 produced from the decomposition of 3.50 moles of NaN3, you can use the stoichiometric ratio between NaN3 and N2.
From the balanced chemical equation:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
The stoichiometric ratio shows that 2 moles of NaN3 produces 3 moles of N2.
Using the ratio, you can set up a proportion to solve for the number of moles of N2 produced:
2 moles of NaN3 / 3 moles of N2 = 3.50 moles of NaN3 / X moles of N2
Cross-multiplying and solving for X gives:
X = (3.50 moles of NaN3 * 3 moles of N2) / 2 moles of NaN3
X = 5.25 moles of N2
Therefore, the decomposition of 3.50 moles of NaN3 produces 5.25 moles of N2.
(b) To find the grams of NaN3 required to form 16.0 g of nitrogen gas, you need to use the molar mass and stoichiometry.
From the balanced chemical equation:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
1 mole of NaN3 has a molar mass of 65.01 g/mol.
Using the stoichiometric ratio, you know that 2 moles of NaN3 produce 3 moles of N2.
First, calculate the number of moles of N2 using the given mass:
16.0 g of N2 / (28.02 g/mol) = 0.571 mol of N2
Using the stoichiometric ratio, calculate the number of moles of NaN3:
0.571 mol N2 * (2 mol NaN3 / 3 mol N2) = 0.381 mol of NaN3
Now, find the mass of NaN3 using the molar mass:
0.381 mol of NaN3 * (65.01 g/mol) = 24.73 g of NaN3
Therefore, 24.73 grams of NaN3 are required to form 16.0 grams of nitrogen gas.
(c) To determine the grams of NaN3 required to produce 19.0 ft3 of nitrogen gas, you need to convert from volume to moles, and then to mass using the given density.
From the balanced chemical equation:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
First, convert the volume of nitrogen gas from ft3 to liters:
19.0 ft3 * (28.32 L/ft3) = 538.08 L
Using the density of 1.25 g/L, you can calculate the mass of N2:
538.08 L * (1.25 g/L) = 672.6 g of N2
Using the stoichiometric ratio, you know that 3 moles of N2 are produced from 2 moles of NaN3.
Calculate the number of moles of N2:
672.6 g of N2 / (28.02 g/mol) = 24.02 mol of N2
Using the stoichiometric ratio, calculate the number of moles of NaN3:
24.02 mol of N2 * (2 mol NaN3 / 3 mol N2) = 16.02 mol of NaN3
Now, find the mass of NaN3 using the molar mass:
16.02 mol of NaN3 * (65.01 g/mol) = 1,041.6 g of NaN3
Therefore, 1,041.6 grams of NaN3 are required to produce 19.0 ft3 of nitrogen gas with a density of 1.25 g/L.
This is a stoichiometry problem. Here is a step by step procedure for working such problems. Note that in c the 19.0 ft^3 must be converted to the metric system in order to use the density in the metric system.
http://www.jiskha.com/science/chemistry/stoichiometry.html