$\text{Bài 10 :}$

$n_{Mg}=2,4/24=0,1mol$

$PTHH :$

$Mg+2HCl\to MgCl_2+H_2↑$

$\text{a.Theo pt :}$

$n_{MgCl_2}=m_{H_2}=n_{Mg}=0,1mol$

$⇒m_{MgCl_2}=0,1.95=9,5g$

$m_{H_2}=0,1.2=0,2g$

$\text{b.Theo pt :}$

$n_{HCl}=2.n_{Mg}=2.0,1=0,2mol$

$⇒m_{HCl}=0,2.36,5=7,3g$

$⇒m_{dd\ HCl}=\dfrac{7,3}{14,6\%}=50g$

$\text{c.Theo ĐLBTKL :}$

$m_{dd\ spu}=m_{Mg}+m_{HCl}-m_{H_2}=2,4+50-0,2=52,2g$

$d.C\%_{MgCl_2}=\dfrac{9,5}{52,2}.100\%=18,2\%$

$\text{Bài 11 :}$

$n_{Zn}=6,5/65=0,1mol$

$n_{H_2SO_4}=1.0,2=0,2mol$

$PTHH :$

                       $Zn  +  H_2SO_4\to ZnSO_4+H_2↑$

$\text{Theo pt :      1 mol      1 mol}$

$\text{Theo đbài : 0,1 mol    0,2 mol}$

Tỷ lệ : $\dfrac{0,1}{1}<\dfrac{0,2}{1}$

$\text{⇒Sau pư H2SO4 dư}$

$\text{a.Theo pt :}$

$n_{H_2SO_4\ pư}=n_{Zn}=0,1mol$

$⇒n_{H_2SO_4\ dư}=0,2-0,1=0,1mol$

$\text{b.Theo pt :}$

$n_{ZnSO_4}=n_{Zn}=0,1mol$

$c.C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M$

Đáp án:

\(10/\\ a,m_{MgCl_2}=9,5\ g.\\ m_{H_2}=0,2\ g.\\ b,m_{ddHCl}=50\ g.\\ c,m_{dd\ spư}=52,2\ g.\\ d,C\%_{MgCl_2}=18,2\%.\\ 11/\\ a,H_2SO_4\ dư.\\ n_{H_2SO_4}(dư)=0,1\ mol.\\ b,n_{ZnSO_4}=0,1\ mol.\\ c,CM_{ZnSO_4}=0,5M.\)

Giải thích các bước giải:

\(10/\\ a,PTHH:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑\\ n_{Mg}=\dfrac{2,4}{24}=0,1\ mol.\\ Theo\ pt:\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\ mol.\\ ⇒m_{MgCl_2}=0,1.95=9,5\ g.\\ ⇒m_{H_2}=0,1.2=0,2\ g.\\ b,Theo\ pt:\ n_{HCl}=2n_{Mg}=0,2\ mol.\\ ⇒m_{ddHCl}=\dfrac{0,2.36,5}{14,6\%}=50\ g.\\ c,BTKL:\\ m_{dd\ spư}=m_{Mg}+m_{ddHCl}-m_{H_2}=2,4+50-0,2=52,2\ g.\\ d,C\%_{MgCl_2}=\dfrac{9,5}{52,2}.100\%=18,2\ %.\) \(11/\\ PTHH:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑\\ n_{Zn}=\dfrac{6,5}{65}=0,1\ mol.\\ Đổi\ 200ml=0,2\ lít.\\ ⇒n_{H_2SO_4}=0,2.1=0,2\ mol.\\ Lập\ tỉ\ lệ:\ \dfrac{0,1}{1}<\dfrac{0,2}{1}\\ ⇒H_2SO_4\ dư.\\ ⇒n_{H_2SO_4}(dư)=0,2-0,1=0,1\ mol.\\ b,Theo\ pt:\ n_{ZnSO_4}=n_{Zn}=0,1\ mol.\\ c,CM_{ZnSO_4}=\dfrac{0,1}{0,2}=0,5M.\)

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