
- A$0$
- ✓$2$
- C$3$
- D$4$

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$4 HNO _{3}(l)+3 KCl ( s ) \rightarrow Cl _{2}( g )+ NOCl ( g )+ 2 H _{2} O ( g )+3 KNO _{3}( s )$
The amount of $HNO _{3}$ required to produce $110.0 \;g$ of $KNO _{3}$ is $...... \;g$
(Given : Atomic masses of $H , O , N$ and $K$ are $1 , 16,14$ and $39$ respectively.)
$[HCl=36.5; {(N{H_4})_2}S{O_4}=132; N{H_3}=17]$
$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n({C_2}{H_4}) \to {( - C{H_2} - C{H_2} - )_n}$
The amount of polyethylene obtained from $64.1\, kg$ $Ca{C_2}$ is......$kg$
[Assume : No cyano complex is formed; $K _{ sp }( AgCN )$ $=2.2 \times 10^{-16}$ and $\left. K _{ a }( HCN )=6.2 \times 10^{-10}\right]$
$\begin{array}{*{20}{c}}
{{H_2}N - CH - CH - CHO} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,} \\
{HOOC\,\,\,\,\,\,\,\,\,\,COOH\,\,\,\,\,}
\end{array}$