In #58*g# of #"ammonium nitrate"#, what mass of nitrogen is present?

1 Answer
anor277
May 31, 2017

Approx. #20*g#.

Explanation:

We work out a molar quantity of #"ammonium nitrate:"#

#"Moles of ammonia nitrate"=(58.0*g)/(80.05*g*mol^-1)=0.725*mol#.

And we take this molar quantity, and multiply it by the appropriately modified molar mass of nitrogen...........

#0.725*molxx2xx14.01*g*mol^-1=??*g#

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