How many moles of nitrogen would react with excess hydrogen to produce 520 ml of ammonia?
a) 0.0116 mol
b) 0.012 mol
c) 0.0232 mol
d) 0.024 mol

n = PV / RT = 1.00(.520 L) / (0.08206 atm L/mol K ) 273 K
n= 0.0232 mol NH3

0.0232 mol NH3 (1 mol N2 / 2 mol NH3) = 0.0116 mol N2

Therefore, the correct answer is A.

3

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d. most models represent a real-world object in a simpler way.

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