Mg+2HCl→MgCl2+H2

Zn+2HCl→ZnCl2+H2

Gọi nMg là a, nZn là b

a=b

→a-b=0

24a+65b=13,35

→a=0,15

   b=0,15

mMgCl=0,15.95=14,25g

mZnCl2=0,15.136=20,4g

m muối=mMgCl2+mZnCl2=14,25+20,4=34,65g

nH2=0,15+0,15=0,3mol

VH2=0,3.22,4=6,72l

H2+CuO→Cu+H2O

nCuO=80/80=1mol

nCuO pứ=nH2=0,3mol

→nCuO dư=1-0,3=0,7mol

mCu=0,3.64=19,2g

mCuO dư=0,7.80=56g

m rắn=mCu+mCuO dư=19,2+56=75,2g

Em tham khảo nha :

\(\begin{array}{l}
1)\\
hh:Mg(a\,mol);Zn(b\,mol)\\
\left\{ \begin{array}{l}
a - b = 0\\
24a + 65b = 13,35
\end{array} \right.\\
 \Rightarrow a = 0,15mol;b = 0,15mol\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,15mol\\
{m_{MgC{l_2}}} = 0,15 \times 95 = 14,25g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,15mol\\
{m_{ZnC{l_2}}} = 0,15 \times 136 = 20,4g\\
{m_m} = 14,25 + 20,4 = 34,65g\\
2)\\
{n_{{H_2}}} = {n_{Mg}} + {n_{Zn}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
3)\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = \dfrac{{80}}{{80}} = 1mol\\
\dfrac{{0,34}}{1} < \dfrac{1}{1} \Rightarrow CuO\text{ dư}\\
{n_{Cu{O_d}}} = 1 - 0,3 = 0,7g\\
{m_{Cu{O_d}}} = 0,7 \times 80 = 56g\\
{n_{Cu}} = {n_{{H_2}}} = 0,3mol\\
{m_{Cu}} = 0,3 \times 64 = 19,2g\\
m = 56+19,2= 75,2g
\end{array}\)