Em tham khảo nha :
10g $HNO_3$ nha em
\(\begin{array}{l}
a)\\
NaOH + HN{O_3} \to NaN{O_3} + {H_2}O\\
{n_{NaOH}} = \dfrac{{10}}{{40}} = 0,25mol\\
{n_{HN{O_3}}} = \dfrac{{10}}{{63}} = 0,159mol\\
\dfrac{{0,25}}{1} > \dfrac{{0,159}}{1} \Rightarrow NaOH\text{ dư}\\
\text{Vậy dung dịch sau phản ứng có tính kiềm}\\
b)\\
{n_{NaO{H_d}}} = 0,25 - 0,159 = 0,091mol\\
{m_{^{NaO{H_d}}}} = 0,091 \times 40 = 3,64g\\
{n_{NaN{O_3}}} = {n_{HN{O_3}}} = 0,159mol\\
{m_{NaN{O_3}}} = 0,159 \times 85 = 13,515g\\
c)\\
NaOH + HN{O_3} \to NaN{O_3} + {H_2}O\\
{n_{HN{O_3}}} = {n_{NaO{H_d}}} = 0,091mol\\
{m_{HN{O_3}}} = 0,091 \times 63 = 5,733g
\end{array}\)