Question

Correct reaction with products are:

• A. $NH_4NO_2 + NaOH + Zn\ powder / \Delta \longrightarrow NH_3 + Na_2ZnO_2$
• B. $NH_4Cl + NaNO_3 \stackrel{\Delta}{\longrightarrow} N_2O + NaCl + H_2O$
• C. $(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 + 4H_2O + Cr_2O_3$
• D. $NH_2CONH_2 \xrightarrow{\frac{NaNO_2}{HCl}} NH_3 + CO_2 \uparrow + NaCl$

Answer 1

Answer: (C)

(iii)

Answer 2

Let's analyze each reaction:

(i) $NH_4NO_2 + NaOH + Zn\ powder / \Delta \longrightarrow NH_3 + Na_2ZnO_2$ This reaction involves the reduction of the nitrite ion ($NO_2^-$) by zinc in an alkaline medium. The nitrogen in $NO_2^-$ (oxidation state +3) is reduced to nitrogen in $NH_3$ (oxidation state -3). The ammonium ion ($NH_4^+$) present in $NH_4NO_2$ will also react with the strong base NaOH to produce $NH_3$. Zinc is oxidized to zincate ($ZnO_2^{2-}$). So, the products $NH_3$ and $Na_2ZnO_2$ are expected. However, the given equation is not stoichiometrically correct. The reduction of $NO_2^-$ to $NH_3$ by Zn in alkaline medium is: $NO_2^- + 3Zn + 7OH^- \longrightarrow NH_3 + 3ZnO_2^{2-} + H_2O$. The reaction of $NH_4^+$ with $OH^-$ is $NH_4^+ + OH^- \longrightarrow NH_3 + H_2O$. Starting with $NH_4NO_2$, the overall reaction would involve both processes. The given equation shows a 1:1:1 ratio of reactants producing 1:1 ratio of products, which is incorrect based on balancing redox and acid-base reactions involved. Therefore, reaction (i) is incorrect as written.

(ii) $NH_4Cl + NaNO_3 \stackrel{\Delta}{\longrightarrow} N_2O + NaCl + H_2O$ Heating a mixture of $NH_4Cl$ and $NaNO_3$ can lead to the formation of $NH_4NO_3$ and $NaCl$ through a double displacement reaction: $NH_4Cl + NaNO_3 \longrightarrow NH_4NO_3 + NaCl$. Ammonium nitrate ($NH_4NO_3$) decomposes upon heating to produce nitrous oxide ($N_2O$) and water: $NH_4NO_3 \stackrel{\Delta}{\longrightarrow} N_2O + 2H_2O$. Combining these, the overall reaction is $NH_4Cl + NaNO_3 \stackrel{\Delta}{\longrightarrow} N_2O + NaCl + 2H_2O$. The given equation shows only one molecule of water as a product, which is incorrect. Therefore, reaction (ii) is incorrect.

(iii) $(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 + 4H_2O + Cr_2O_3$ This is the thermal decomposition of ammonium dichromate. It is a well-known reaction, often demonstrated as a "volcano" experiment. Let's check if the equation is balanced. Reactants: N = 2, H = 8, Cr = 2, O = 7. Products: N in $N_2$ = 2, H in $4H_2O$ = $4 \times 2 = 8$, Cr in $Cr_2O_3$ = 2, O in $4H_2O + Cr_2O_3$ = $4 \times 1 + 3 = 7$. The equation is balanced, and the products are correctly identified. This reaction is correct.

(iv) $NH_2CONH_2 \xrightarrow{\frac{NaNO_2}{HCl}} NH_3 + CO_2 \uparrow + NaCl$ This reaction involves urea ($NH_2CONH_2$) reacting with sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$). $NaNO_2$ and $HCl$ react to form nitrous acid ($HNO_2$). Urea reacts with nitrous acid to produce nitrogen gas ($N_2$), carbon dioxide ($CO_2$), and water ($H_2O$): $NH_2CONH_2 + 2HNO_2 \longrightarrow N_2 \uparrow + CO_2 \uparrow + 3H_2O$. Since $HNO_2$ is formed from $NaNO_2 + HCl$, the overall reaction is: $NH_2CONH_2 + 2(NaNO_2 + HCl) \longrightarrow N_2 \uparrow + CO_2 \uparrow + 3H_2O + 2NaCl$. The given equation shows ammonia ($NH_3$) as a product instead of nitrogen gas ($N_2$). Therefore, reaction (iv) is incorrect.

Based on the analysis, only reaction (iii) is correctly represented with the correct products and stoichiometry.