`a) Mg+2HCl->MgCl_2+H_2`

`b)`

`n_[Mg]=[4.8]/24=0.2` (mol)

PT: `n_[MgCl_2]=n_[Mg]=0.2` (mol)

`-> m_[MgCl_2]=0.2 * 95=19` (g)

`c)`

PT: `n_[HCl]=2n_[Mg]=2*0.2=0.4` (mol)

`V_[dd HCl]=[0.4]/2=0.2 (L)=200(mL)`

`d)`

PT: `n_[H_2]=n_[Mg]=0.2` (mol)

`V_[H_2] = 0.2 * 24.79=4.958(L)`

Lời giải:

$\rm n_{Mg}=\dfrac{4,8}{24}=0,2\ (mol)$

$\rm a)$

Phản ứng xảy ra:

$\rm Mg+2HCl\to MgCl_2+H_2\\\kern0 cm0,2\kern,05cm\to\kern,1cm0,4\kern,2cm\to\kern,2cm0,2\kern,05cm\to\kern,1cm0,2\kern1cm(mol)$

$\rm b)\\m_{MgCl_2}=0,2\cdot 95=19\ (g)\\c)\\V_{HCl}=\dfrac{0,4}{2}=0,2\ (L)\\d)\\V_{H_2}=0,2\cdot 24,79=4,958\ (L)$