Em tham khảo nha :

\(\begin{array}{l}
a)\\
BaC{l_2} + N{a_2}C{O_3} \to BaC{O_3} + 2NaCl\\
{m_{BaC{l_2}}} = \dfrac{{208 \times 10}}{{100}} = 20,8g\\
{n_{BaC{l_2}}} = \dfrac{{20,8}}{{208}} = 0,1mol\\
{n_{N{a_2}C{O_3}}} = {n_{BaC{l_2}}} = 0,1mol\\
{C_{{M_{N{a_2}C{O_3}}}}} = \dfrac{{0,1}}{{0,1}} = 1M\\
b)\\
{n_{BaC{O_3}}} = {n_{BaC{l_2}}} = 0,1mol\\
{m_{BaC{O_3}}} = 0,1 \times 197 = 19,7g\\
c)\\
{n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
{m_{NaCl}} = 0,2 \times 58,5 = 11,7g\\
{m_{{\rm{dd}}N{a_2}C{O_3}}} = 100 \times 1,05 = 105g\\
C{\% _{NaCl}} = \dfrac{{11,7}}{{208 + 105 - 19,7}} \times 100\%  = 3,99\% \\
d)\\
{C_{{M_{NaCl}}}} = \dfrac{{0,2}}{{0,1}} = 2M
\end{array}\)

\(\begin{array}{l}
2)\\
a)\\
Fe + CuS{O_4} \to Cu + FeS{O_4}\\
{n_{Cu}} = \dfrac{{3,2}}{{64}} = 0,05mol\\
{n_{Fe}} = {n_{Cu}} = 0,05mol\\
{m_{Fe}} = 0,05 \times 56 = 2,8g\\
b)\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,05mol\\
{C_{{M_{CuS{O_4}}}}} = \dfrac{{0,05}}{{0,05}} = 1M\\
c)\\
{n_{FeS{O_4}}} = {n_{Cu}} = 0,05mol\\
{m_{FeS{O_4}}} = 0,05 \times 152 = 7,6g\\
{m_{ddCuS{O_4}}} = 50 \times 1,6 = 80g\\
C{\% _{FeS{O_4}}} = \dfrac{{7,6}}{{80 + 2,8 - 3,2}} \times 100\%  = 9,55\% \\
3)\\
a)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}} = 0,05 \times 2 = 0,1mol\\
{n_{NaOH}} = {n_{HCl}} = 0,1mol\\
{m_{NaOH}} = 0,1 \times 40 = 4g\\
C{\% _{NaOH}} = \dfrac{4}{{50}} \times 100\%  = 8\% \\
b)\\
{n_{NaCl}} = {n_{HCl}} = 0,1mol\\
{m_{NaCl}} = 0,1 \times 58,5 = 5,85g\\
{m_{ddHCl}} = 50 \times 1,18 = 59g\\
C{\% _{NaCl}} = \dfrac{{5,85}}{{50 + 59}} \times 100\%  = 5,37\% \\
c)\\
{C_{{M_{NaCl}}}} = \dfrac{{0,1}}{{0,05}} = 2M
\end{array}\)