6/
$PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol.$
$⇒V_{H_2}=0,3.22,4=6,72l.$
$c,Theo$ $pt:$ $n_{HCl}=3n_{Al}=0,6mol.$
$⇒m_{HCl}=0,6.36,5=21,9g.$
$⇒m_{ddHCl}=\dfrac{21,9}{14,6\%}=150g.$
$d,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{32}{80}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,4}{1}>\dfrac{0,3}{1}$
$⇒Cu$ $dư.$
⇒ Chất rắn sau pư gồm $CuO$ (dư) và $Cu$
$⇒n_{CuO}(dư)=0,4-\dfrac{0,3.1}{1}=0,1mol.$
$⇒m_{CuO}(dư)=0,1.80=8g.$
$Theo$ $pt:$ $n_{Cu}=n_{H_2}=0,3mol.$
$⇒m_{Cu}=0,3.64=19,2g.$
8/
$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$n_{Zn}=\dfrac{19,5}{65}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,3mol.$
$⇒V_{H_2}=0,3.22,4=6,72l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Zn}=0,4mol.$
$⇒V_{HCl}=\dfrac{0,4}{1}=0,4l=400ml.$
$c,PTPƯ:2A+2H_2O\xrightarrow{t^o} 2AOH+H_2$
$Theo$ $pt:$ $n_{A}=2n_{H_2}=0,6mol.$
$⇒M_{A}=\dfrac{13,8}{0,6}=23g/mol.$
⇒ A là Natri (Na)
9/
$a,PTPƯ:Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3mol.$
$Theo$ $pt:$ $n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol.$
$⇒m_{Fe}=0,2.56=11,2g.$
$b,PTPƯ:Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$
$Theo$ $pt:$ $n_{H_2SO_4}=n_{Fe}=0,2mol.$
$⇒m_{H_2SO_4}=0,2.98=19,6g.$
$⇒m_{ddH_2SO_4}=\dfrac{19,6}{19,6\%}=100g.$
$c,Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,2mol.$
$⇒m_{H_2}=0,2.2=0,4g.$
$⇒m_{dd spư}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}$
$⇒m_{dd spư}=11,2+100-0,4=109,8g.$
$Theo$ $pt:$ $n_{FeSO_4}=n_{Fe}=0,2mol.$
$⇒m_{FeSO_4}=0,2.152=30,4g.$
$d,C\%_{H_2SO_4}=\dfrac{19,6}{100+50}.100\%=13,06\%$
chúc bạn học tốt!
