Na3N decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced
how many moles of Na3N was used? (22.4 L = 1 mole of any gas)
2Na3N --> 6Na + N2

Answers

Answer 1

Answer: 1.224 moles of [tex]Na_3N[/tex] were used.

Explanation:

We are given:

Volume of nitrogen gas produced = 13.7 L

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas

Applying unitary method:

13.7 L of nitrogen gas will be occupied by = [tex]\frac{1mol}{22.4L}\times 13.7L=0.612mol[/tex]

For the given chemical reaction:

[tex]2Na_3N\rightarrow 6Na+N_2[/tex]

By Stoichiometry of the reaction:

1 mole of nitrogen gas is produced by 2 mole of [tex]Na_3N[/tex]

So, 0.612 moles of nitrogen gas will be produced from = [tex]\frac{2}{1}\times 0.612=1.224mol[/tex] of [tex]Na_3N[/tex]

Hence, 1.224 moles of [tex]Na_3N[/tex] were used.


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Answers

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Answers

Answer:

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2Na + Cl2 ---> 2NaCl

Using the molar masses , the theoretical yield is:

46 g of Na yields  46 + 35.5*2 g NaCl

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[tex]{\huge{\boxed{\sf{\pink{☛Answer࿐}}}}}[/tex]

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HERE'S YOUR ANSWER

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