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Question

Sodium azide (NaN3) is an explosive chemical used in automobile airbags. It is made by the following reaction:
NaNO3+3NaNH2NaN3+3NaOH+NH3

One takes 17.0 g of NaNO3 and 13.0 g of NaNH2. If this amount is used in automobile bags, how much N2 is formed due to collision at STP?

A
3.73 L
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B
12.93 L
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C
8.62 L
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D
2.15 L
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Solution

The correct option is A 3.73 L
NaNO31 mol17 g+3NaNH23 mol13 gNaN3+3NaOH+NH3

Finding the limiting reagent,

moles of NaNO3=1785=0.2 mol moles of NaNH2=1339=13 mol

moles of NaNO31=0.21
moles of NaNH23=13×3

Hence, NaNH2 is the limiting reactant.

Amount of NaN3 is obtained from NaNH2,
3 moles NaNH2 give = 1 mole NaN3
13 mole gives =13×3mol NaN3=19 mole NaN3

2NaN33N2

2 moles NaN3 gives = 3 mole N2 gas

19 mole NaN3 give =32×9mole N2=16 mole N2
1 mole N2 at STP = 22.4 L
16 mole N2 at STP = 3.73 L

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