Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: 2NaN3(s)→2Na(s)+3N2(g)

a)How many moles of N2 are produced by the decomposition of 1.60mol of NaN3?
b)How many grams of NaN3 are required to form 14.0g of nitrogen gas?
c)How many grams of NaN3 are required to produce 10.0ft3 of nitrogen gas if the gas has a density of 1.25 g/L?

a) 2.4 moles

a) To find the number of moles of N2 produced by the decomposition of 1.60 moles of NaN3, we can use the mole ratio from the balanced equation: 2 mol NaN3 produces 3 mol N2.

Therefore, using the mole ratio:
1.60 mol NaN3 * (3 mol N2 / 2 mol NaN3) = 2.40 mol N2

So, 1.60 moles of NaN3 will produce 2.40 moles of N2.

b) To determine the grams of NaN3 required to form 14.0 grams of nitrogen gas, we need to convert the mass of N2 to moles and then use the mole ratio from the balanced equation.

First, we convert the mass of N2 to moles:
14.0 g N2 * (1 mol N2 / 28.01 g N2) = 0.499 mol N2

Now, using the mole ratio:
0.499 mol N2 * (2 mol NaN3 / 3 mol N2) * (65.01 g NaN3 / 1 mol NaN3) = 21.7 g NaN3

Therefore, 14.0 grams of nitrogen gas requires 21.7 grams of NaN3.

c) To determine the grams of NaN3 required to produce 10.0 ft^3 of nitrogen gas, we first need to convert the volume of nitrogen gas to liters.

10.0 ft^3 * (28.32 L / 1 ft^3) = 283.2 L

Next, we can calculate the amount of moles of N2 using the ideal gas law: PV = nRT, assuming standard conditions (STP) where the pressure is 1 atm and the temperature is 273 K.

n = PV / RT
= (1 atm) * (283.2 L) / (0.0821 atm·L/mol·K * 273 K)
= 11.02 moles of N2

Using the mole ratio:
11.02 mol N2 * (2 mol NaN3 / 3 mol N2) * (65.01 g NaN3 / 1 mol NaN3) = 476 g NaN3

So, 10.0 ft^3 of nitrogen gas with a density of 1.25 g/L requires 476 grams of NaN3.

a) To determine the number of moles of N2 produced by the decomposition of 1.60 mol of NaN3, we need to use the stoichiometric ratio from the balanced equation: 2NaN3(s) → 2Na(s) + 3N2(g).

According to the balanced equation, 2 moles of NaN3 produce 3 moles of N2.

Given that you have 1.60 mol of NaN3, you can set up a simple ratio to find the number of moles of N2:

(3 mol N2 / 2 mol NaN3) * 1.60 mol NaN3 = 2.40 mol N2

Therefore, 1.60 mol of NaN3 would produce 2.40 mol of N2.

b) To calculate the grams of NaN3 required to form 14.0 g of nitrogen gas, you need to use the stoichiometric ratio and the molar mass of NaN3.

First, calculate the molar mass of NaN3:
Na has a molar mass of 22.99 g/mol, and N has a molar mass of 14.01 g/mol. Since there are three nitrogen atoms in NaN3, the molar mass of NaN3 is:

(3 * 14.01 g/mol) + 22.99 g/mol = 65.00 g/mol.

Next, use the stoichiometric ratio to find the moles of NaN3 required to produce 14.0 g of N2:

(2 mol NaN3 / 3 mol N2) * (14.0 g N2 / 28.01 g/mol) = 4.67 mol NaN3.

Finally, convert the moles of NaN3 to grams:

4.67 mol NaN3 * 65.00 g/mol = 303.11 g NaN3.

Therefore, you would need 303.11 grams of NaN3 to produce 14.0 grams of nitrogen gas.

c) To find the mass of NaN3 required to produce 10.0 ft3 of nitrogen gas at a density of 1.25 g/L, you need to convert the volume to liters.

Since 1 ft3 is approximately equal to 28.32 liters, we can find the volume in liters:

(10.0 ft3 / 1) * (28.32 L / 1 ft3) = 283.2 L.

Now, calculate the number of moles of N2 using the ideal gas law:

PV = nRT.

The pressure (P) is 1 atm, the temperature (T) is constant, and the gas constant (R) is 0.0821 L.atm/mol.K.

So, n (moles of N2) = PV / RT = (1 atm * 283.2 L) / (0.0821 L.atm/mol.K * T).

Since the temperature is not given, we can't determine the exact number of moles of N2. However, assuming typical room temperature (25°C or 298 K), we can calculate the number of moles.

n = (1 atm * 283.2 L) / (0.0821 L.atm/mol.K * 298 K) = 11.62 moles of N2.

Finally, use the stoichiometric ratio to find the moles of NaN3 required:

(2 mol NaN3 / 3 mol N2) * 11.62 mol N2 = 7.74 mol NaN3.

Now, convert the moles of NaN3 to grams:

7.74 mol NaN3 * 65.00 g/mol = 503.10 g NaN3.

Therefore, you would need 503.10 grams of NaN3 to produce 10.0 ft3 of nitrogen gas with a density of 1.25 g/L.

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