Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{m_{HCl}} = \dfrac{{300 \times 3,65}}{{100}} = 10,95g\\
b)\\
{n_{HCl}} = \dfrac{{10,95}}{{36,5}} = 0,3mol\\
{n_{ZnO}} = \dfrac{{8,1}}{{81}} = 0,1mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,3}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = 0,3 - 0,1 \times 2 = 0,1mol\\
{m_{HC{l_d}}} = 0,1 \times 36,5 = 3,65g\\
{n_{ZnC{l_2}}} = {n_{ZnO}} = 0,1mol\\
{m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g\\
C{\% _{ZnC{l_2}}} = \dfrac{{13,6}}{{8,1 + 300}} \times 100\%  = 4,41\% \\
C{\% _{HCl}} = \dfrac{{3,65}}{{8,1 + 300}} \times 100 = 1,18\% \\
2)\\
{m_{CuS{O_4}.5{H_2}O}} = 50g\\
 \Rightarrow {m_{CuS{O_4}}} = \frac{{50 \times 160}}{{250}} = 32g\\
{m_{{H_2}O}} = 50 - 32 = 18g\\
{m_{dd}} = 700 + 50 = 750g\\
C{\% _{CuS{O_4}}} = \dfrac{{32}}{{750}} \times 100\%  = 4,267\% 
\end{array}\)

1.

ZnO+2HCl→ZnCl2+H2O

nZnO=8,1/81=0,1mol

mHCl=300.3,65/100=10,95g

nHCl=10,95/36,5=0,3mol

→nHCl dư=0,3-0,1.2=0,1mol

mdd sau pứ=8,1+300=308,1g

mZnCl2=0,1.136=13,6g

C%ZnCl2=13,6/308,1.100%=4,41%

mHCl dư=0,1.36,5=3,65g

C%HCl dư=3,65/308,1.100%=1,18%

2.

nCuSO4.5H2O=50/250=0,2mol

→nCuSO4=0,2mol

mCuSO4=0,2.160=32g

mdd sau pứ=50+700=750g

C%CuSO4=32/750.100%=4,26%