Under standard conditions, what volume does a #25*g# mass of ammonia gas occupy?

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Answer 1 · 2017-02-28T05:39:14

Approx. #55*L#

#"Moles of ammonia "= (25.0*g)/(17.01*g*mol^-1)=1.47*mol#.

Given the stoichiometry of the equation, I need #3/2# equiv dihydrogen gas, i.e. #2.20*mol#.

I don't know what your syllabus says in regard to the molar volume of an ideal gas at STP; the standard molar volume is #24.8*L*mol^-1#, so we get:

#2.20*molxx24.8*L*mol^-1=??L#