1.
BTKL:
$n_{HCl}=\dfrac{15-10}{36,5}=\dfrac{10}{73}(mol)$
Amin đơn chức ($RNH_2$) nên $n_X=n_{HCl}=\dfrac{10}{73}(mol)$
$\to M_X=\dfrac{10}{\dfrac{10}{73}}=73$
$\to M_R=73-16=57$ ($C_4H_9$)
$\to$ CT amin là $C_4H_9NH_2$
Đồng phân cấu tạo:
$CH_3-CH_2-CH_2-CH_2NH_2$
$CH_3-CH(C_2H_5)-NH_2$
$CH_3-CH(CH_3)-CH_2NH_2$
$(CH_3)_3CNH_2$
$CH_3-CH_2-NH-CH_2-CH_3$
$CH_3-NH-CH_2-CH_2-CH_3$
$CH_3-NH-CH(CH_3)-CH_3$
$CH_3-N(CH_3)-CH_2-CH_3$
$\to 8$ CTCT
2.
Bảo toàn khối lượng:
$n_{HCl}=\dfrac{23,76-15}{36,5}=0,24(mol)$
$\to V=\dfrac{0,24}{0,75}.1000=320ml$