Câu 4 :

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

$n_{HCl}=18,25/36,5=0,5mol$

$a.$                    $Zn  +  2HCl\to ZnCl_2+H_2↑$

b.Theo pt :    1 mol      2 mol

Theo đbài :  0,3 mol   0,5 mol

Ta có tỷ lệ :

$\dfrac{0,3}{1}>\dfrac{0,5}{2}$

⇒Sau pư Zn dư , HCl phản ứng hết

Theo pt :

$n_{Zn\ pư}=1/2.n_{HCl}=1/2.0,5=0,25mol$

$⇒n_{Zn\ dư}=0,3-0,25=0,05mol$

c.Theo pt :

$n_{H_2}=1/2.n_{HCl}=1/2.0,5=0,25mol$

$⇒V_{H_2}=0,25.22,4=5,6l$

Câu 5 :

$n_{Na}=4,6/23=0,2mol$

$a.2Na+2H_2O\to 2NaOH+H_2↑$

b.Theo pt :

$n_{H_2}=1/2.n_{Na}=1/2.0,2=0,1mol$

$⇒V_{H_2}=0,1.22,4=2,24l$

c.Theo pt :

$n_{NaOH}=n_{Na}=0,2mol$

$⇒m_{NaOH}=0,2.40=8g$

$d.m_{dd\ spu}=4,6+200-0,1.2=204,4g$

$⇒C\%_{NaOH}=\dfrac{8}{204,4}.100\%≈3,91\%$

 

4/

$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$

$b,n_{Zn}=\dfrac{19,5}{65}=0,3mol.$

$n_{HCl}=\dfrac{18,25}{36,5}=0,5mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,3}{1}>\dfrac{0,5}{2}$

$⇒Zn$ $dư.$

$⇒n_{Zn}(dư)=0,3-\dfrac{0,5.1}{2}=0,05mol.$

$⇒m_{Zn}(dư)=0,05.65=3,25g.$

$c,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,25mol.$

$⇒V_{H_2}=0,25.22,4=5,6l.$

5/

$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$

$b,n_{Na}=\dfrac{4,6}{23}=0,2mol.$

$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1mol.$

$⇒V_{H_2}=0,1.22,4=2,24l.$

$c,Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,2mol.$

$⇒m_{NaOH}=0,2.40=8g.$

$d,m_{ddNaOH}=4,6+200-(0,1.2)=204,4g.$

$⇒C\%_{NaOH}=\dfrac{8}{204,4}.100\%=3,91\%$

chúc bạn học tốt!