Câu 1:
$NH_2-CH_2-COOH+NaOH\to NH_2-CH_2-COONa + H_2O$
$n_{\text{gly}}=\dfrac{7,5}{75}=0,1(mol)$
$n_{NaOH}=0,1(mol)$
$\Rightarrow$ phản ứng vừa đủ.
$n_{NH_2-CH_2-COONa}=0,1(mol)$
$\Rightarrow m_{\text{muối}}=0,1.97=9,7g$
Câu 2: (sửa "ankin" thành "alanin)
a,
$NH_2-CH(CH_3)-COOH + HCl\to ClNH_3-CH(CH_3)-COOH$
b,
$n_{HCl}=0,2.0,15=0,03(mol)$
$\Rightarrow n_{\text{Ala}}=0,03(mol)$
$\Rightarrow m=0,03.89=2,67g$