In the reaction #NH_4NO_2(s) -> N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP?

Answer 1

If #50*L# of water gas were evolved, then we can immediately see that a #25*L# volume of dinitrogen was likewise evolved.

I follow the reaction as written:

#NH_4NO_2(s) rarr N_2(g) + 2H_2O(g)#.

Both products are gaseous. Because #Vpropn# at constant temperature and pressure, the stoichiometry is easy to assess. Of course if #50*L# of #H_2O(l)# were evolved, it would be a different story.

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Answer 2

Natalie Anton

According to the stoichiometry of the reaction, for every mole of NH4NO2 that decomposes, one mole of N2 gas is produced. Since the molar volume of an ideal gas at STP (standard temperature and pressure) is 22.4 L/mol, and the reaction produces 2 moles of H2O gas for every mole of N2 gas, the volume of N2 gas produced can be calculated using the stoichiometry of the reaction.

Given that 50.0 L of H2O gas is produced, and the ratio of N2 to H2O is 1:2, the volume of N2 gas produced is: (50.0 L H2O) / (2 mol H2O) * (1 mol N2 / 1 mol NH4NO2) * (22.4 L/mol) = 560 L N2 gas.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

In the reaction #NH_4NO_2(s) -&gt; N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP?

In the reaction #NH_4NO_2(s) -> N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP?