Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
{C_{{M_{CuS{O_4}}}}} = \dfrac{{0,3}}{{0,2}} = 1,5M\\
b)\\
{n_{NaOH}} = \dfrac{{16}}{{40}} = 0,4mol\\
{C_{{M_{NaOH}}}} = \dfrac{{0,4}}{{0,2}} = 2M\\
c)\\
{n_{HCl}} = 0,2 \times 2 + 0,3 \times 5 = 1,9mol\\
{C_{{M_{HCl}}}} = \dfrac{{1,9}}{{0,2 + 0,3}} = 3,8M\\
d)\\
{n_{NaCl}} = 0,3 \times 2 = 0,6mol\\
{V_{{H_2}O}} = 200ml\\
{C_{{M_{NaCl}}}} = \dfrac{{0,6}}{{0,2 + 0,3}} = 1,2M\\
2)\\
a)\\
{n_{HCl}} = 0,2 \times 2,5 = 0,5mol\\
b)\\
{m_{HCl}} = \dfrac{{200 \times 7,3}}{{100}} = 14,6g\\
{n_{HCl}} = \dfrac{{14,6}}{{36,5}} = 0,4mol\\
c)\\
{m_{NaOH}} = \dfrac{{300 \times 40}}{{100}} = 120g\\
{n_{NaOH}} = \dfrac{{120}}{{40}} = 3mol\\
d)\\
{n_{NaOH}} = 0,5 \times 0,5 = 0,25mol\\
3)\\
a)\\
C\% = \dfrac{{30}}{{30 + 170}} \times 100\% = 15\% \\
b)\\
{m_{NaCl}} = 30 + \dfrac{{170 \times 20}}{{100}} = 64g\\
C\% = \dfrac{{64}}{{30 + 170}} \times 100\% = 32\% \\
c)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{9,3}}{{62}} = 0,15mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,3mol\\
{m_{NaOH}} = 0,3 \times 40 = 12g\\
C\% = \dfrac{{12}}{{9,3 + 190,7}} \times 100\% = 6\% \\
d)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = \dfrac{{4,6}}{{23}} = 0,2mol\\
{n_{NaOH}} = {n_{Na}} = 0,2mol\\
{m_{NaOH}} = 0,2 \times 40 = 8g\\
{m_{ddspu}} = 4,6 + 195,6 - 0,1 \times 2 = 200g\\
C\% = \dfrac{8}{{200}} \times 100\% = 4\% \\
4)\\
a)\\
{m_{NaCl}} = \dfrac{{200 \times 0,9}}{{100}} = 1,8g\\
{m_{{H_2}O}} = 200 - 1,8 = 198,2g\\
b)\\
{n_{NaOH}} = 0,5 \times 0,5 = 0,25mol\\
{V_{{H_2}O}} = 500ml = 0,5l
\end{array}\)
