bài 1:

a) $Zn$ + $H_{2}$$S$$O_{4}$ → ZnS$O_{4}$ + $H_{2}$$↑$

b) $n_{Zn}$ =$\frac{m}{M}$ =$\frac{9,1 }{65}$ =$0,14$ $mol$

$Zn$ + $H_{2}$$S$$O_{4}$ →ZnS$O_{4}$  + $H_{2}$$↑$

theo pt ta có:

$n_{H_{2}}$ =$n_{Zn}$=$0,14$ $mol$

⇒$V_{H_{2}}$ $=n.22,4=0,14.22,4=3,136l$

c) $n_{H_{2}SO_{4}}$ =$n_{Zn}$=$0,14$ $mol$

$m_{ZnSO_{4} }$ $=n.M=0,14.161=22.54g$

bài 2:

a) $Zn$ +2 $HCl$ →$Zn$$Cl_{2}$ + $H_{2}$$↑$

b) $n_{H_{2}}$=$\frac{V}{22,4}$ =$\frac{3,36}{22,4}$ $=0,15 mol$

$Zn$ + 2$HCl$ → $Zn$$Cl_{2}$ + $H_{2}$$↑$

theo pt ta có:

$n_{HCl}$ =$\frac{0,15.2}{1}$ $=0,3 mol$

⇒$m_{HCl}$ $=n.M=0,3.36,5=10,95g$

c) $H_{2}$ + $CuO$ →($t^{o}$trên mũi tên) $Cu$ + $H_{2}$$O$

xin hay nhất 

1)

Phản ứng xảy ra:

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{9,1}}{{65}} = 0,14{\text{ mol}} = {n_{ZnS{O_4}}} = {n_{{H_2}}}\)

\( \to {V_{{H_2}}} = 0,14.22,4 = 3,136{\text{ lít}}\)

\({m_{ZnS{O_4}}} = 0,14.(65 + 96) = 22,54{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)

\( \to {n_{HCl}} = 2{n_{{H_2}}} = 0,15.2 = 0,3{\text{ mol}}\)

\( \to {m_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}}\)

Dẫn khí \(H_2\) qua \(CuO\)

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

\({n_{Cu}} = {n_{{H_2}}} = 0,15{\text{ mol}}\)

\( \to {m_{Cu}} = 0,15.64 = 9,6{\text{ gam}}\)