How many grams of #N_2# are produced from the reaction of 8.96 g of #H_2O_2# and 6.68 g of #N_2H_4#?

1 Answer
Aug 9, 2018

Answer:

Approx. #4*g#...

Explanation:

Hydrazine is OXIDIZED to dinitrogen gas...

#H_2N-NH_2 rarr N_2+4H^++ 4e^(-)# #(i)#

The nitrogens in hydrazine have a #-II# oxidation state...

Hydrogen peroxide is REDUCED to water...

#HO-OH+2H^+ +2e^(-)rarr 2H_2O # #(ii)#

Note that when we got element-element bonds, they are conceived to share the bonding electrons. The carbons in ethane have formal oxidation states of #C(-III)#... The oxygens in hydrogen peroxide have formal oxidation states of #O(-I)#...

And so we take #(i)+2xx(ii)# to retire the electrons....

#H_2N-NH_2 +2HO-OHrarr N_2+4H_2O#

And so now we assess the molar quantities of each reagent...

#"Moles of hydrogen peroxide"=(8.96*g)/(34.01*g*mol^-1)=0.264*mol#

#"Moles of hydrazine"=(6.68*g)/(32.05*g*mol^-1)=0.208*mol#

And thus hydrogen peroxide is the reagent in MOLAR deficiency (why they could not have added excess is beyond me!)….

We could get at most...#0.264*molxx1/2xx28.01*g*mol^-1=??*g#