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!

Câu 6 :

$n_{Al}=5,4/27=0,2mol$

$a/2Al+6HCl\to 2AlCl_3+3H_2$

b/Theo pt :

$n_{H_2}=3/2.n_{Al}=3/2.0,2=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

c/Theo pt :

$n_{HCl pư}=3.n_{Al}=3.0,2=0,6mol$

$⇒m_{HCl pư}=0,6.36,5=21,9g$

$⇒m_{ddHCl}=\dfrac{21,9}{14,6\%}=150g$

Câu 8 :

$n_{Zn}=19,5/65=0,3mol$

$PTHH :$

$Zn+2HCl\to ZnCl_2+H_2↑$

a/Theo pt :

$n_{H_2}=n_{Zn}=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

b/Theo pt :

$n_{HCl}=2.n_{Zn}=2.0,3=0,6mol$

$⇒V_{ddHCl}=\dfrac{0,6}{1}=0,6l$

$c/PTPU :$

$2A+2H_2O\to 2AOH+H_2↑$

Theo pt :

$n_A=2.n_{H_2}=2.0,3=0,6mol$

$⇒M_A=\dfrac{13,8}{0,6}=23(Na)$

Vậy A là Na

Câu 9 :

$n_{H_2}=6,72/22,4=0,3mol$

$PTHH:$

$Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O$

a/Theo pt :

$n_{Fe}=2/3.n_{H_2}=2/3.0,3=0,2mol$

$⇒m_{Fe}=0,1.56=11,2g$

$b/PTHH :$

$Fe+H_2SO_4\to FeSO_4+H_2↑$

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 :

$n_{H_2}=n_{FeSO_4}=n_{Fe}=0,2mol$

$⇒m_{FeSO_4}=0,2.152=30,4g$

$m_{ddspu}=11,2+100-0,2.2=110,8g$

$⇒C\%_{FeSO_4}=\dfrac{30,4}{110,8}.100\%=27,44\%$

$d/C\%_{H_2SO_4}=\dfrac{19,6}{100+50}.100\%≈13,06\%$