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------------------------------------------------ At STP, how many liters of [tex]NH_3[/tex] can be produced from the reaction of 6.00 mol of [tex]N_2[/tex] with 6.00 mol of [tex]H_2[/tex]?

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

at STP, 267.47 liters of NH3 can be produced from the reaction of 6.00 mol of N2 with 6.00 mol of H2.
Using the balanced chemical equation, we see that 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3. Therefore, with 6.00 mol of N2 and 6.00 mol of H2, we have enough reactants to produce:

(6.00 mol N2) x (2 mol NH3 / 1 mol N2) = 12.00 mol NH3

Now we can use the ideal gas law to find the volume of NH3 at STP (standard temperature and pressure):

PV = nRT

At STP, T = 273 K and P = 1 atm. We can assume that the volume of the reactants and products are all the same (since they are all gases), so we can use the same volume for NH3 as we would for N2 and H2.

V = (nRT) / P

V = (12.00 mol NH3) x (0.0821 L atm / mol K) x (273 K) / (1 atm)

V = 267.47 L NH3

Therefore, at STP, 267.47 liters of NH3 can be produced from the reaction of 6.00 mol of N2 with 6.00 mol of H2.

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