$a)C\%=\frac{7,3.100}{7,3+92,7}=7,3\% ; C_M=\frac{7,3.1000}{36,5.92,7}\approx 2,16 M$$b)H_2SO_4+NaCl\rightarrow NaHSO_4+HCl ....................!?:$Thực tế !
$H_2SO_4+2NaCl\rightarrow Na_2SO_4+2NaCl$
$\Rightarrow \begin{cases}n_{NaCl}=0,2 mol \\n_{H_2SO_4}=0,1 mol \end{cases}\Rightarrow \begin{cases}m_{NaCl}=11,7 gam \\m_{H_2SO_4 98\%}=10 gam \end{cases}$
$c)NaOH+HCl\rightarrow NaCl+H_2O$
$\begin{cases}n_{HCl}=0,2 mol \\n_{NaOH}=0,4 mol \end{cases}\Rightarrow n_{NaOH}>n_{HCl}\Rightarrow pH>7:$dung dịch có tính bazo
$\begin{cases}\%NaOH=\frac{0,2.40.100}{100+160}=3,08\% \\\%NaCl=\frac{0,2.58,5.100}{100+160}=4,5\% \end{cases}$
$d)AgNO_3+HCl\rightarrow AgCl\downarrow+HNO_3$
$n_{AgNO_3}=0,2 mol\Rightarrow V=\frac{0,2}{30,5} lit$