# 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?

Aug 9, 2018

Approx. $4 \cdot g$...

#### Explanation:

Hydrazine is OXIDIZED to dinitrogen gas...

${H}_{2} N - N {H}_{2} \rightarrow {N}_{2} + 4 {H}^{+} + 4 {e}^{-}$ $\left(i\right)$

The nitrogens in hydrazine have a $- I I$ oxidation state...

Hydrogen peroxide is REDUCED to water...

$H O - O H + 2 {H}^{+} + 2 {e}^{-} \rightarrow 2 {H}_{2} O$ $\left(i i\right)$

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 $C \left(- I I I\right)$... The oxygens in hydrogen peroxide have formal oxidation states of $O \left(- I\right)$...

And so we take $\left(i\right) + 2 \times \left(i i\right)$ to retire the electrons....

${H}_{2} N - N {H}_{2} + 2 H O - O H \rightarrow {N}_{2} + 4 {H}_{2} O$

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

$\text{Moles of hydrogen peroxide} = \frac{8.96 \cdot g}{34.01 \cdot g \cdot m o {l}^{-} 1} = 0.264 \cdot m o l$

$\text{Moles of hydrazine} = \frac{6.68 \cdot g}{32.05 \cdot g \cdot m o {l}^{-} 1} = 0.208 \cdot m o l$

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