$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{l})+2 \mathrm{~N}_{2} \mathrm{H}_{4}(\mathrm{l}) \rightarrow 3 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ hydrarine

Which substance acts as the reducing agent?
$\overline{\mathrm{N}_{2} \mathrm{H}_{4}}$

Question

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{l})+2 \mathrm{~N}_{2} \mathrm{H}_{4}(\mathrm{l}) \rightarrow 3 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ hydrarine

Which substance acts as the reducing agent?
$\overline{\mathrm{N}_{2} \mathrm{H}_{4}}$

Accepted Answer

***D*** ∻Key Concept∻ ⚹Reducing Agent⚹ ∻Explanation∻ ⚹In the given reaction, $\mathrm{N_2H_4}$ (hydrazine) is the reducing agent because it donates electrons to $\mathrm{N_2O_4}$, resulting in its reduction to $\mathrm{N_2}$, while $\mathrm{N_2H_4}$ itself gets oxidized to $\mathrm{H_2O}$.⚹◊What is a reducing agent and how does it function in a chemical reaction?⍭ Generate me a similar question◊

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