How many liters of nitrogen gas are needed to react with 90.0 g of potassium at STP in order to produce potassium nitride according to the following reaction? 6 K + N 2 ---> 2 K 3 N

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Answer 1 · 2018-03-23T17:32:43

Let us calculate the molar quantity of metal...but we should first write the stoichiometric equation to inform our calculations..

...we gots...

#3K(s) + 1/2N_2(g) rarr K_3N(s)#

#"Moles of potassium"=(90.0*g)/(39.1*g*mol^-1)=2.30*mol#.

And we require #1/6*"equiv dinitrogen"#, i.e. #0.384*mol#.

Now the molar volume at #"STP"-=22.4*L* mol^-1#...at least it is according to my syllabus.

And so we take the product...

#22.4*L* mol^-1xx0.384*mol~=9*L#.