Question

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`?

Answer 1

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`