Sodium azide (NaN3) can decompose at 300 o C to produce sodium metal (Na) and ni

Question

Sodium azide (NaN3) can decompose at 300 o C to produce sodium metal (Na) and nitrogen gas (N2). This is one of the key reactions in an automobile air bag. The balanced equation for this reaction is: 2NaN3(s) ?2Na (s) + 3 N2(g)

a.How many g of NaN3 would be needed to produce 2.26 L of nitrogen gas at the same temperature and pressure? Use the correct number of significant figures in your answer.

b. How many g of NaN3 would be need to produce 1.00 quart of nitrogen gas at the same temperature and pressure? (1.057 quart = 1.00 L) Enter your answer with the correct number of significant figures.

Explanation / Answer

a) According to the stoichiometric balanced equation we get

2NaN3 ----- 2Na + 3N2

From this equation we can observe that

2 moles of sodium azide produces 3 moles of nitrogen gas.

At 1 atm volume of 1 mole of gas should be 22.4 litres.

Hence, volume of 3 moles of nitrogen equals 3*22.4 = 67.2 litres.

2.26 litres of nirogen is produced from 2/67.2*2.26 = 0.0336 moles of sodium azide.

1 mole of sodium azide = 65.00987 grams.

ANSWER a. Hence, 0.0336 moles of sodium azide equals 2.184 grams of sodium azide.

b) 1 quart equals 0.946 litres.

2/67.2*0.946 = 0.0282 moles of sodium azide produces 1 quart of nitrogen gas.

Therefore, 0.0282 moles = 1.833 grams of sodium azide.

Hence, Answer b) 1.833 grams of sodium azide.

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