What mass of oxygen is required to make #90*g# of #NO#?

1 Answer
Jan 26, 2017

Answer:

Approx. #42*g#

Explanation:

We could assume either atoms or molecules; for this problem it is wise to use atomic mass.

#"Atoms of oxygen"-=(48*g)/(16.00*g*mol^-1)=3.0*mol*"oxygen atoms, i.e "3xxN_A" oxygen atoms."#

So we need #3.0*mol# #"nitrogen atoms"#, or #3xxN_A# #"nitrogen atoms,"# which has an equivalent mass of,

#3*molxx14.01*g*mol^-1=??*g#