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Thermal decomposition of sodium azide (NaN₃) into nitrogen gas and sodium metal is used to inflate car airbags. The reaction proceeds by the following chemical reaction.
2NaN₃(s)⟶3N₂(g)+2Na(s)
How many grams of NaN₃ are need to inflate a 10.00L airbag with nitrogen gas at 1.000 atm and 273.15 K assuming ideal gas behavior?

Respuesta :

Answer:

19.5g

Explanation:

Firstly, we can use the ideal gas equation to calculate the number of moles of nitrogen gas needed.

We use the following formula:

PV = nRT

n = PV/RT

In the question, we were given:

P = 1 atm

V = 10L

T = 273.15k

R = molar gas constant= 0.082L.atm/mol.k

Let us insert these values into the ideal gas equation.

n = (1 * 10)/(273.15 * 0.082) = 0.45moles

The number of moles of nitrogen required is 0.45 moles

From the balanced reaction equation, we can see that 2 moles of sodium azide yielded 3 moles of Nitrogen.

Hence the number of moles of sodium azide produced from 0.45 moles of nitrogen would be : (0.45 * 2)/3 = 0.3mole

To get the mass of sodium azide needed, we simply multiply the number of moles of sodium azide by the molar mass of sodium azide. We know the number of moles but we do not know the molar mass. The molar mass of sodium azide is 23 + 3(14) = 23 + 42 = 65g/mol

The mass = 65 * 0.3 = 19.5g

pstnonsonjoku

The mass of azide required is 19.3 g.

The equation of the reaction is given as; 2NaN₃(s)⟶3N₂(g)+2Na(s)

We have to obtain the number of moles of the nitrogen gas from;

P = 1.000 atm

V = 10.00L

T= 273.15 K

R = 0.082 atm L K-1mol-1

n = PV/RT

n =  1.000 atm × 10.00L/0.082 atm L K-1mol-1 × 273.15 K

n = 0.446 moles

Now;

2 moles of azide yields 3 moles of nitrogen

x moles of azide yieilds 0.446 moles of nitrogen

x = 2 moles × 0.446 moles/3 moles

x = 0.297 moles

Mass of azide required=  0.297 moles × 65 g/mol = 19.3 g

Learn more about sodium azide: https://brainly.com/question/17960050

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