End product formed in the above reaction is

$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_3}(excess) + B{r_2}\xrightarrow{{hv}}}
\end{array}$ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,Br}
\end{array}}\limits_{(A)} $ + $\mathop {\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_2} - Br}
\end{array}}\limits_{(B)} $

the percentage yields of the products $(A)$ and $(B)$ are expected to be

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The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $\mathrm{NiCl}_2 .6 \mathrm{H}_2 \mathrm{O}$ to form a stable coordination compound. Assume that both the reactions are $100 \%$ complete. If $1584 \mathrm{~g}$ of ammonium sulphate and $952 \mathrm{~g}$ of $\mathrm{NiCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}$ are used in the preparation, the combined weight (in grams) of gypsum and the nickelammonia coordination compound thus produced is $\qquad$ (Atomic weights in $\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \overline{\mathrm{Cl}=35.5}, \mathrm{Ca}=40, \mathrm{Ni}=59$ )

→

The major product obtained when $B{r_2}/Fe$ is treated with is