16.934 grams of water is formed when 12.5 liter NH3 (at 298 k and 1.50 atm) is reacted with 18.9 l of o2 (at 323 k and 1.1 atm).
Using the formula of ideal gas equation PV = RT
Where P is pressure, V is volume, R is Avogadro’s constant and T is temperature
We calculate the given as,
number of NH3 = n1
n1= PV/RT
= 0.7663 moles
number of moles of O2 = n2
n2 = 0.784 moles
now, - 5 moles of O2-------------6 moles of H20
hence, 0.784 moles will give
6 x 0.784 / 5 = 0.941 moles of H2O
0.941 moles of H2O = 16.934 gms
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