$n_{Mg}=5,6/24≈0,23mol$

$a.$                  $Mg  +  2HCl\to MgCl_2+H_2↑$

b.Theo pt :     1 mol      2 mol

Theo đbài :   0,23 mol   0, 4 mol

Ta có tỷ lệ :

$\dfrac{0,23}{1}>\dfrac{0,4}{2}$

⇒Sau pư Mg dư , HCl pư hết

Theo pt :

$n_{H_2}=1/2.n_{HCl}=1/2.0,4=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$

c.Theo pt :

$n_{MgCl_2}=1/2.n_{HCl}=1/2.0,4=0,2mol$

$⇒m_{MgCl_2}=0,2.95=19g$

$m_{dd\ spu}=5,6+100-0,2.2=105,2g$

$⇒C\%_{MgCl_2}=\dfrac{19}{105,2}.100\%=18,06\%$

 

$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$

$b,n_{Mg}=\dfrac{5,6}{24}=\dfrac{7}{30}mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{\frac{7}{30}}{1}>\dfrac{0,4}{2}$

$\text{⇒ Mg dư.}$

$\text{⇒ Tính theo}$ $n_{HCl}$

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

$⇒V_{H_2}=0,2.22,4=4,48l.$

$c,Theo$ $pt:$ $n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,2mol.$

$⇒m_{MgCl_2}=0,2.95=19g.$

$⇒m_{dd\ spư}=m_{Mg}+m_{ddHCl}-m_{H_2}=5,6+100-(0,2.2)=105,2g.$

$⇒C\%_{MgCl_2}=\dfrac{19}{105,2}.100\%=18,06\%$

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