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:

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

Under standard conditions, what volume does a 25*g25*g mass of ammonia gas occupy?