Đáp án:
Giải thích các bước giải:
$n_{Zn}=$ $\dfrac{26}{65}=0,4(mol)$
$Zn+2HCl→ZnCl_2+H_2$
$0,4→$ $0,8→$ $0,4$ $→0,4$ $(mol)$
a. $V_{H_2}=0,4.22,4=8,96(l)$
b. $m_{HCl}=0,8.36.,5=29,2(g)$
c. $m_{ZnCl_2}=0,4.136=54,4(g)$
$n_{CuO}=$ $\dfrac{8}{80}=0,1(mol)$
$ CuO+H_2\xrightarrow{t^o}Cu+H_2O$
$Bđ$ $0,1$ $0,4$
$Pư$ $0,1→$$0,1→$ $0,1$ $(mol)$
$Spư$ $0$ $0,3$
$m_{Cu}=0,1.64=6,4(g)$